DFT shift theorem proof

Question

Learning DSP on my own time. Can't figure out the proof for DFT shift theorem which states the following:

Given, $x[n]$ to be a periodic with period $N$, $\text{DFT}\{x[n]\} = X[k]$, then $$ DFT\{x[n-a]\} = e^{-j\frac{2\pi}{N}a}X[k] $$

I found a proof here, but I can't figure out how did they leap from $$\sum_{m = -\Delta}^{N-1-\Delta}{e^{-j\frac{2\pi}{N}k(m+\Delta)}x[m]}$$ to $$ \sum_{m = 0}^{N-1}{e^{-j\frac{2\pi}{N}km} e^{-j\frac{2\pi}{N}k\Delta} x[m]} $$

Specifically how did they change the limits of the summation without changing the variable?

Answer 1

For an $N$-periodic sequence $y[n]$, the sum over one period is always the same, no matter how you choose the first index:

$$\sum_{n=0}^{N-1}y[n]=\sum_{n=\Delta}^{N-1+\Delta}y[n]$$

for any integer $\Delta$. Try to verify that the sequence you sum over is indeed $N$-periodic.