Euler's theorem for powers of 2 According to Euler's theorem,
$$x^{\varphi({2^k})} \equiv 1 \mod 2^k$$
for each $k>0$ and each odd $x$. Obviously, number of positive integers less than or equal to $2^k$ that are relatively prime to $2^k$ is 
$$\varphi({2^k}) = 2^{k-1}$$
so it follows that
$$x^{{2^{k-1}}} \equiv 1 \mod 2^k$$
This is fine, but it seems like even
$$x^{{2^{k-2}}} \equiv 1 \mod 2^k$$
holds, at least my computer didn't find any counterexample.
Can you prove or disprove it?

We can look at this from another view.    
Let's first rewrite the question as: is it true, that $2^k | x^{2^{k-2}}-1 $?    
Next we recall, that $\small x^2-1 = (x+1)(x-1)$ and thus $\small x^{2^a}-1=(x^{2^{a-1}}+1)(x^{2^{a-1}}-1)$      
So we can now rewrite the question for some top-exponent $ \small b=k-2$
$$ M_b=x^{2^b}-1 = (x^{2^{b-1}}+1)(x^{2^{b-2}}+1)(x^{2^{b-3}}+1)\cdots(x^{2^0}+1)(x^{2^0}-1)$$    
This are b+1 factors by simply counting of parentheses. Now x must be odd such that each parenthese is even. Then each parenthese contains at least the primefactor 2 to the power of 1, so we know already: at least $2^{b+1} $ divides $M_b$.   
But, the odd $x$ is either $4y+1$ or $4y+3$, thus one of the two last factors, $(x+1)$ or $(x-1)$ must be divisible by $2^2$ because they form either $\small ((4y+1)+1)((4y+1)-1)$ or $\small ((4y+3)+1)((4y+3)-1)$.     
So one of those two parentheses adds (at least) one more primefactor 2, and it follows that (at least) $2^{b+2}$ divides $M_b$ and we have, that indeed
$$2^k | (x^ {2^{k-2}}-1 )$$
