Why do we get a reflected electromagnetic wave when it hits a perfect conductor? We've an EM wave

$\vec{E_i}=\vec{E_0}e^{i(\omega t-kz)}$

As it reaches on the surface of the perfect conductor we know the electric field must be zero, so we deduce that another electric field must be produced which will cancel the original field on the surface.
So we can quickly say that the induced field is equal to the inverted incident field on the surface.
But we cannot say that a reflected wave is produced from this logic.
We can only say that an induced field exists, which is opposite to that of the incident one and lives on the surface of the conductor.
And it doesn't follow that this induced field should should travel out of the surface in the form of a reflected wave.
How can then one deduce that a reflected EM wave exists when a EM wave strikes a perfect conductor?
Thank you.
 A: It is true that the electric field inside a perfect conductor is zero.
But consider what is happening on the surface of the conductor.

We can only say that an induced field exists, which is opposite to that of the incident one and lives on the surface of the conduit conductor.

The incident electromagnetic wave moves the free charges on the conductor which produces a current that then creates a radiating field which is the reflected wave.

We can only say that an induced field exists, which is opposite to that of the incident one and lives on the surface of the conduit conductor.

And this field is oscillating charges on the surface of the conductor.

And it doesn't follow that this induced field should should travel out of the surface in the form of a reflected wave.

It is these induced (changing) fields on the surface that then cause the reflected electromagnetic wave.

How can then one deduce that a reflected EM wave exists when a EM wave strikes a perfect conductor?

Since the electric field inside the conductor is zero, there is an infinite impedance to the electromagnetic wave right at the surface. Now if we take the phase of the electromagnetic wave at this interface to be $0$ degrees (and since the conductor allows no electric field), there must be an electromagnetic wave with an opposite phase of $180$ degrees to cancel the incident electromagnetic wave at the interface.
A: The presence of a reflected wave is simply a consequence of Maxwell's equations and the boundary conditions imposed on their solutions.
When a wave is incident upon the conducting interface, you are free to try any solution you like for what happens to the electromagnetic fields on either side of the interface. But those fields must be solutions to Maxwell's equations in (I) a good conductor on one side of the interface and (II) vacuum (or whatever) on the incidence side of the interface. Secondly, the components of the electric fields tangential to the interface must be continuous.
The very low impedance in the conductor means that the electric field is almost zero in the conductor, which means it must be almost zero on the incidence side too.
So what solution could we explore for the fields on the incidence side. If we treat the fields as the sum of the incident wave plus some other, unknown, solution to Maxwell's equations, then our options are limited.
We know that the solutions to Maxwell's equations in vacuum (or whatever dielectric it is) must be electromagnetic waves. We know that to maintain the electric field at zero close to the interface, it must have the same frequency and amplitude as the incident wave. We know that this wave must transport almost all the energy in the incident wave away from the interface. And we know that this energy does not travel into the conductor. Finally, we know that this wave must travel at a specific angle (equal to the angle of incidence), because otherwise it would not cancel with the incident wave all along the interface. All that we are left with is a wave of equal amplitude and frequency travelling in the opposite direction (for normal incidence).
A: The simplest explanation is the following. For a perfect conductor the incoming light makes the conduction electrons move 180 degrees out of phase with it. These electrons create two waves that are 180 degrees out of phase with the incoming field. One is moving along with the incoming wave and cancels it beyond the surface of the metal. The other moves opposite to it and is the reflected wave. This is what Maxwell's equations say if the so called bound charge is used as the source of the field. So Maxwell in vacuum but with the conduction current as a source.
Alternatively the bound charge can be absorbed in an effective medium, characterized by a non-unity relative dielectric constant. To find the answer with this approach you need to use the Fresnel equations.
OP states that "it doesn't follow that this induced field should should travel out of the surface". This follows from Maxwell's equations.
A: My first answer gives a more "holistic" approach, just arguing on the basis of having valid solutions to Maxwell's equations. However, here is another approach that looks in detail at the co-existing fields and current densities, using the microscopic version of Ohm's law. This demonstrates that the reflected wave is indeed produced by currents near the surface of the conductor and that the fields produced by these currents also cancel out the incident electric field further into the conductor.
My initial disquiet about this explanation was based on the fact that for good conductors, the E-field inside the conductor is proportional to $\sigma^{-1/2}$, where $\sigma$ is the conductivity, implying that the current density was proportional to $\sigma^{1/2}$. Thus despite the possibility of $\sigma$ varying by many orders of magnitude in different (good) conductors, the reflected wave amplitude is always very close to the incident wave amplitude.
The key to resolving this is to realise that as the conductivity gets higher, the electric field in the conductor gets lower, but the resultant current density multiplied by the skin depth is almost constant and exactly that required to produced the reflected (and transmitted) waves.
Take as a set-up and incident wave of the form $\vec{E_i} = E_i \cos(\omega t-kz)\hat{i}$, which is incident normally on a conductive plane at $z=0$, with a high conductivity $\sigma$. Let us assume this leads to a transmitted wave in the conductor and a reflected wave. Let us further assume that the wavelength of the incident field is much larger than any skin depth of the conductor, so that in the region where there is a current density the value of $\vec{E_i} \simeq \vec{E_i}(0,t)$.
If the average current density in the medium is $\vec{J}(\omega t)$ and the effective width of the current carrying region is $\delta$, then the electric field generated by the oscillating current sheet will be$\dagger$
$$ \vec{E_s}(z,t) = -\frac{\mu_0 c \delta}{2} \vec{J}(\omega[t- z/c]) \ \ \ \ \ {\rm for\ \ z>0}$$
$$ \vec{E_s}(z,t) = -\frac{\mu_0 c \delta}{2} \vec{J}(\omega[t+ z/c]) \ \ \ \ \ {\rm for\ \ z<0}$$
where $t - z/c$ represents a retarded time.
The total electric fields in the regions $z>0$ and $z<0$ will be $\vec{E_T} = \vec{E_i} + \vec{E_s}$ and from Ohm's law $\vec{J}(t) = \sigma \vec{E_T}(0,t)$.
At $z=0$, we can use either expression for the field generated by the oscillating current sheet to write
$$\vec{J}(t) = \sigma \left( E_i \cos(\omega t)\hat{i} - \frac{\mu_0 c \delta}{2}\vec{J}(t)\right) $$
$$ \vec{J}(t) = \frac{\sigma E_i\cos(\omega t)}{1 + \sigma \mu_0 c \delta/2}\hat{i}\ . $$
This current density can then be used to find $\vec{E_s}$ at $z>0$ and $z<0$.
$$ \vec{E_s} = -\frac{\mu_0 c \delta}{2} \left( \frac{\sigma E_i\cos(\omega t - kz)}{1 + \sigma \mu_0 c \delta/2}      \right)\hat{i} \ \ \ \ \ {\rm for\ \ z>0}$$
$$ \vec{E_s} = -\frac{\mu_0 c \delta}{2} \left( \frac{\sigma E_i\cos(\omega t + kz)}{1 + \sigma \mu_0 c \delta/2}      \right)\hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$
These fields add to the incident wave to give the total electric field on either side of the current sheet:
$$ \vec{E_T} = E_i \cos(\omega t - kz) \left[ 1 - \frac{\sigma \mu_0 c \delta}{2 + \sigma \mu_0 c \delta}        \right] \hat{i}  \ \ \ \ \ {\rm for\ \ z>0}$$
$$ \vec{E_T} = E_i \left[ \cos(\omega t - kz) - \frac{\sigma \mu_0 c \delta}{2 + \sigma \mu_0 c \delta}\cos(\omega t + kz)  \right] \hat{i}  \ \ \ \ \ {\rm for\ \ z<0}$$
For a very good conductor we can require that $\sigma \mu_0 c \delta \gg 2$, these reduce to
$$\vec{E_T} \simeq 0\ \ \ \ \ {\rm for\ \ z>0}$$
$$\vec{E_T} \simeq 2E_i \sin(kz)\sin(\omega t) \hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$
Thus the fields produced by the oscillating current sheet are exactly those required to cancel the incident electric field on the far side of the conductive boundary (and beyond the current-carrying region) and to produce a standing wave on the incident side of the boundary.
I am still thinking about whether I can refine this treatment to deal with the spatial dependence of the electric fields and current density within the current-carrying region rather than assuming it is narrow.
The solution to the paradox I posed in my initial remarks is that although the current density in the conductor is $\propto \sigma^{1/2}$, what matters for the fields that are produced is the product $J \delta$; and since the skin depth is $\propto \sigma^{-1/2}$, this product is almost constant. This is why the reflected wave has almost the same amplitude, independent of the exact value of $\sigma$ (as long as it is large enough to be considered a good conductor).
$\dagger$ Proving the form of the fields radiating from an oscillating current sheet.
Assume a position $\vec{r}=z\hat{k}$ relative to an  origin at some point on the plane $z=0$ and that there is a time-dependent current density $J(t) = J\cos(\omega t) \hat{i}$ in a sheet of thickness $\delta$.
The general solution of the inhomogeneous wave equation is
$$\vec{A}(z,t) = \frac{\mu_0}{4\pi} \int \frac{J\cos[\omega(t-|r-r'|/c)]}{|r-r'|}\ d^3r'\ \hat{i},$$
where $\vec{r'}$ is a position on the current sheet, $|r-r'| = \sqrt{z^2 + r'^2}$. If we also let $\vec{J}\ d^3r' = \delta \vec{J} 2\pi r'\ dr'$ and the solution for the vector potential becomes
$$\vec{A}(z,t) = \frac{\mu_0 \delta J}{4\pi} \int^{\infty}_{r'=0} \frac{\cos[\omega(t - k\sqrt{z^2 + r'^2})]}{\sqrt{z^2 + r'^2}}\ 2\pi r'\ dr'\ \hat{i} .$$
If we let $u = k\sqrt{z^2 + r'^2}$, then $du = k r'\ dr'/\sqrt{z^2 + r'^2}$ and
$$\vec{A}(z,t) = \frac{\mu_0 \delta J}{2k} \int^{\infty}_{u=k|z|} \cos(\omega t - u)\ du\ \hat{i} $$
$$\vec{A}(z,t) = -\frac{\mu_0 \delta J}{2 k} \left[\sin(\omega t-k|z|) - \sin (\infty)\right]\ \hat{i}\ . $$
Taking the time derivative
$$ \vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\frac{\mu_0 \delta J \omega}{2 k}\cos(\omega t - kz)\ \hat{i}\ \ \ \ {\rm for}\ z>0 $$
$$\vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\frac{\mu_0 \delta J \omega}{2 k}\cos(\omega t + kz)\ \hat{i}\ \ \ \ {\rm for}\ z<0\, , $$
where $\omega/k = c$ in vacuum.
