You understand this in one dimension and then three dimensions is obvious.
Consider a function $f(x)$.
If we define the Fourier transform of this function by
$$ \tilde{f}(k) = \int \, dx \, f(x) e^{- i k x}$$
then the original function $f(x)$ is can be written as
$$ f(x) = \int \frac{dk}{2\pi} \tilde{f}(k) e^{i k x} \, .$$
The $2 \pi$ is there because the equation simply is not true without it.
To prove this you really have to use strict analytical mathematics.
Anyway, the point is that if you want to expand a function of position in terms of $\exp[i k x]$ functions (i.e. plane waves) then you need the $2\pi$ in the reverse transform.
If we had defined the Fourier transform like this
$$ \tilde{f}(\nu) = \int dx \, f(x) e^{- i 2 \pi \nu x}$$
then the inverse transform would be
$$ f(x) = \int d\nu \, \tilde{f}(\nu) e^{i 2 \pi \nu x} \, .$$
Therefore you see that you either need to put the $2\pi$ in both of the exponentials or leave it out of the exponentials but divide the reverse transform by $2\pi$.$^{[a]}$
These two sets of transformations are just related by a change of variables $\nu = k/2\pi$, so you can just remember one of them and easily recover the other.
Since the wave number $k$ is defined as $k \equiv 2 \pi / \lambda$ that means that $\nu = 1 / \lambda$.
Physicists tend to prefer the first pair of transformations because they don't like writing the $2 \pi$ in the argument of the trig functions.
In other words, they prefer things like $\cos(kx)$ instead of $\cos(2 \pi x / \lambda)$.
You see this in solid state physics where wave functions are expanded in terms of functions like $\exp[i \vec{k} \cdot \vec{x}]$, and this is the root of why you're seeing factors of $2\pi$ in the equations that go back and forth between position space and $k$ space.
$[a]$: Actually there are even more options but the two I showed are the most common ones used in physics and engineering.