I want to solve the following question from the Digital Image Processing by Gonzalez and Woods book:
Show that the DFT of the discrete function $f(x,y)=1$ is: $$ \Im\left\{ 1\right\} =\hat{h}(u,v)=\begin{cases} 1 & u=v=0\\ 0 & \text{otherwise} \end{cases} $$
What I did is to use the definition: $$ \hat{h}(u,v)=\sum_{x=0}^{M-1}\sum_{y=0}^{N-1}h(x,y)\exp\left(-2\pi iux\frac{1}{M}\right)\exp\left(-2\pi ivy\frac{1}{N}\right) $$ We know that $h(x,y)=1$ so we get: $$ \hat{h}(u,v)=\sum_{x=0}^{M-1}\sum_{y=0}^{N-1}\exp\left(-2\pi iux\frac{1}{M}\right)\exp\left(-2\pi ivy\frac{1}{N}\right) $$ Now I'm stuck and not sure how to progress from here. Do I need to split into cases? For $u=v=0$? How to continue?
Hint
Observe that $$ \sum_{x=0}^{M-1}\sum_{y=0}^{N-1}a_xb_y= (\sum_{x=0}^{M-1}a_x) (\sum_{y=0}^{N-1}b_y) $$ and $$ \sum_{x=0}^{M-1}\exp(cx)=\frac{\exp(cM)-1}{\exp(c)-1}\quad,\quad \exp(c)\ne 1 $$