Why does an electron shell further away from nucleus has higher energy level? Using electrical potential energy $V=\frac{1}{4\pi \varepsilon_0} \frac{Q_1 Q_2}{r}$ , a particle further away from nucleus has lower magnitude of energy.
Using Coulomb's law, a particle further away from nucleus experiences weaker attraction, hence less energy is needed to maintain orbit$^\star$ around that e-shell compared to a electron shell closer to nucleus, hence the one closer to nucleus supposedly should have higher energy.
$^\star$I know in reality $e^-$ does not orbit around a atom, but its position exists as a probability density or radial probability function.
 A: The energy in a level $n$ is given by
$$E  = - \frac{Z^2 R_E}{n^2} $$
where $R_E$ is the Rydberg energy ($R_E = 13.6\mathrm{eV}$).
Therefore, greater $n$ means lower energy (in absolute value), i.e., the electron is less bounded.
A: The potential energy stored in a two like charge system will increase with decrease in distance between them. While for a two unlike charge system, the potential energy decreases with decrease in distance (means potential energy gets liberated if they come close), accounting for increase in attraction.  
In the equation, you provided, the potential energy in the nucleus-electron system is negative. This means the potential energy of the system is liberated and hence indicate attraction of the nucleus with the electron (this is how they attain stability).   
Hence a system comprising of an electron far off from the nucleus will have high potential energy stored in it, indicating they have sufficient potential energy that can overcome the attractive forces (means the attractive forces between the electron and the nucleus is less). This means the potential energy liberated by an electron far from the nucleus is very less. Hence the outermost electrons are less stable.   
For an electron very close to the nucleus, the potential energy is minimum, which means the system comprising of nucleus and a nearer electron liberates most of it's potential energy (so that the system will now have a lesser potential energy) to have an increased attractive force, which in turn corresponds to maximum stability.   
So a large amount of energy is required to liberate an electron from an inner most shell rather than an electron from the outermost shell. This is why we say that the electron in the outermost shell has a higher (potential) energy than the inner most shells. So a less amount of energy is needed to liberate the electron from the outermost shell.  
A: By
$$E=−\frac{Z^2R_E}{n^2}$$
where $R_E$ is the Rydberg energy
As n increase, $E_{PE}$ becomes less -ve(i.e. more +ve) , indicating higher energy level
Or
$$E_{PE} =- \frac{( Q_{proton} Q_{e-})} {4\pi \epsilon r}$$
As r increase, $E_{PE}$ becomes less -ve (i.e. more +ve), indicating a higher energy level
Thanks to everyone that helped!
A: 
By E=−Z^2RE/n2 where RE is the Rydberg energy As n increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level<

Or

EPE = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level<
Thanks to everyone that helped !<

I beg to differ on the above explanation provided by the author @ De Day:
The highest energy acquired by an electron is at K shell , and slowly energy decreases as one moves to L,M,N  ...shells.
the confirmation is the energy required to take out a K-shell electron is highest and in X-ray emission the  high speed cathode electrons knock out K-shell electrons and it needs about 20-25 keV of energy .
Therefore I wish to add that energy levels which are closest to the nucleus is at the highest and the above contention by the author is not correct.
Moreover if a K-shell electron is knocked out and a vacancy is created then any transition from L.M....levels leads to emission lines of lowest wavelength and highest frequency X-rays characteristic lines .
This energy packet contains the difference of energy levels of the atom.
the magnitude of this energy also suggests that the E(K)-E(m)= h. frequency . is largest.
I think the confusion is that the bound states total energy is sum of its  K.E. and P.E. and total energy has to be negative for bound states and it's highest as one moves closer to the nuclear charge +ze.
By just thinking about the potential energy, one has to consider that the nuclear charge field has done work on the electron to get it to a shell radius and this work done is highest if one goes closer.
the test is to supply energy to pull out a K-electron and the  value of energy needed to extract  will again be largeer  than L, M,.. and other shell electrons.
