<s>[INST] Prove that $5^{2^{m-3}}\equiv 1+2^{m-1} \bmod 2^m$. Prove that $5^{2^{m-3}}\equiv 1+2^{m-1} \bmod 2^m$.
The hints given are induction and binomial theorem.
For $m=4$, $5^2=25\equiv 9 \bmod 2^4$ where $9=1+2^3$.
So the base case is done.  
Suppose $5^{2^{m-3}}\equiv 1+2^{m-1} \bmod 2^m$.
Then $5^{2^{m-2}}=5\cdot 5^{2^{m-3}} \equiv 5(1+2^{m-1}) \bmod 2^m $.
Write $5=1+4$. So the term becomes $1+2^{m-1}+2^2+2^{m-2}$.
I am not sure how the binomial theorem can be applied here.  
 [/INST] 
$5^{2^{m-2}}=5\cdot 5^{2^{m-3}}$

This is incorrect. We have $5^{2^{m-2}}=5^{2^{m-3}\times 2}=(5^{2^{m-3}})^2$.
Inductive step : Supposing that $5^{2^{m-3}}=1+2^{m-1}+k\cdot 2^m$ where $k\in\mathbb Z$ gives
$$\begin{align}5^{2^{m-2}}&=(5^{2^{m-3}})^2\\\\&=(1+2^{m-1}+k\cdot 2^m)^2\\\\&=1+2^{2m-2}+k^2\cdot 2^{2m}+2^m+k\cdot 2^{m+1}+k\cdot 2^{2m}\\\\&=1+2^m+(2^{m-3}+k^2\cdot 2^{m-1}+k+k\cdot 2^{m-1})\cdot 2^{m+1}\\\\&\equiv 1+2^m\pmod{2^{m+1}}\end{align}$$
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