I am a high-school math student. I found a pattern a while ago - given n, an odd positive integer, and r, any positive integer power of two, n^r modulo 2r always equals 1. I was trying to find a proof of this, but I could not find anything. I am asking whether there is any way to get any progress on this problem, or a theorem that states this.
Edit - The problem is solved using Euler's theorem or via an induction argument.
Apart from Euler's theorem, it can also be proved straightforwardly by induction on the power of 2.
To prove: $(2k+1)^{(2^r)} \equiv 1 \mod {2^{r+1}}$ for all $r\in\mathbb{N}$.
This is obviously true for $r=0$, because $2k+1\equiv 1 \bmod 2$.
Suppose if is true for a particular value of $r$. Then:
$$(2k+1)^{(2^r)} = 1+m\cdot 2^{r+1}$$
Square both sides:
$$(2k+1)^{(2^{r+1})} = (1+m\cdot 2^{r+1})^2\\
(2k+1)^{(2^{r+1})} = 1+2m\cdot 2^{r+1} + m^2 2^{2r+2}\\
(2k+1)^{(2^{r+1})} = 1+(m+m^2 2^r)\cdot 2^{r+2}\\
(2k+1)^{(2^{r+1})} \equiv 1 \mod 2^{r+2}$$
which means it is also true for $r+1$.
Therefore by induction the statement is true for all values $r\in\mathbb{N}$.
