2-adic valuation of $5^{2^{k-2}}-1$ I have the following exercice :

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*Consider that the 2-adic valuation of $x$ is $k \geq 2$. Prove that the 2-adic valuation of $(1+x)^2- 1$ is $k+1$.


*Using the previous question, conclude that for $k\geq 2$, the 2-adic valuation of $5^{2^{k-2}}-1$ is $k$.
For the first l question, we know that $(1+x)^2 - 1= x(x+2) $. So, $$v_2((1+x)^2-1) = v_2(x)+\min(v_2(x),v_2(2))= k+1.$$
For the last question, I tried several methods but didn't work. Can someone give me a hint please?
Thanks!

*

*If $v_2(x)=k\ge 2$ then $v_2(x+2)=1$ and $v_2( (1+x)^2-1)=v_2 (x(x+2)) =v_2 (x)+v_2(x+2) =k+1$.


*Let $x_{k+2}=5^{2^k}-1=(1+x_{k+1})^2-1, v_2(x_2)=2$, by induction if $v_2(x_k)=k$ then $v_2(x_{k+1})=k+1$.
