Independence of Noise at Each DFT Output
Let $m=k-k'$. Then the sum in your final equation becomes

$$\sum_{n=0}^{N-1}e^{-2\pi imn/N}=\frac{1-e^{-2\pi im}}{1-e^{-2\pi im/N}},\quad m\neq 0\tag{1}$$

where I've used the formula for the geometric series. Noting that $e^{-2\pi im}=1$ for any integer $m$ shows that the sum in $(1)$ equals zero.

[Note that your sums should only have $N$ elements, i.e., their indices should go from $0$ to $N-1$ (instead of $N$).]