DFT and IDFT deriviation problems
What you derive in your answer is that the iDFT of the DFT of a signal is the signal itself. The missing step is that $$ \sum_{k=0}^{N-1}\exp(j2\pi uk/N)=\sum_{m\in\mathbb{Z}}N\delta[u-mN].$$ Then, you get the identity $f[n] = \sum_m f[m]\delta[n-m]$.

However, this does not help for your problem. You need to express the IDFT in terms of DFT operations. So, you need to get

$$
X[m] = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}x[k]\exp(j2\pi km/N)
$$

from a function, which can only calculate

$$
A[m] = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}a[k]\exp(-j2\pi km/N).
$$

If you conjugate both sides of the first equation, you get:

$$
X^*[m]=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}x[k]^*\exp(-j2\pi km/N).
$$

This equation is of the structure of the second equation, when you substitute $A[m]=X^*[m]$ and $a[k]=x^*[k]$. Hence, if you perform the DFT the conjugated signal, you get the conjugated of the IDFT of signal:

$$DFT(X^*)=IDFT(X)^*.$$

So, conjugating both sides yields
$$DFT(X^*)^*=IDFT(X)$$
which is exactly what you need to do to solve your problem: 


Perform elementwise conjugation of the signal
Perform DFT of the signal
Perform elementwise conjugation of the result


Then, you get back the IDFT of the signal. It's analoguous in the 2D-case (i.e. images). Also, note that you might need to consider a different normalization factor (which would correspond to elementwise division by a scalar), depending on your definition of the DFT. If it is with $\frac{1}{\sqrt{N}}$, then normalization is OK as it is. 

Edit: As Royi has pointed out, if your original signal is real, you can omit the 2nd conjugation (since the signal is already real after the DFT, and conjugation won't do anything). As images are most often real-valued, this simplication is very likely to hold here.