Proving modular statements

Question

Show that for any even number $k$, $$2^{nk} \equiv 1 \pmod {2^n + 1} $$

I tried using induction with base case $k = 0$, but I am unsure how to proceed.

Answer 1

Let $$2^n \equiv −1\, {(\mod2^n+1)}\implies {(2^n)}^k \equiv -1^k \pmod {2^n + 1}\implies$$ $$2^{nk} \equiv (-1)^k \pmod{2^n + 1} $$ Therefore $$ 2^{nk} \equiv 1 \pmod {2^n + 1} \iff k := \{2x\,|\, x\in \mathbb{Z}\}$$

Answer 2

$2^{n}$ is congruent to $(-1)$ modulo $2^n+1$.

So $2^{nk}$ is congruent to $(-1)^k=1$ because $k$ is even.