The title wasn't a spoiler at all . So we have $v_p(2^{2^n}+1)=k.$ To show that $v_p(2^{p-1}-1)=k$ If $p=2,$ then $v_2(2^{2^n}+1)=0=v_2(2^{p-1}-1).$ If $p$ is odd, then we know that $2^{p-1}\equiv 1 \mod p.$ We are given that $k\ge 1.$ So $p|2^{2^n}+1.$ Hence $$p\nmid 2^{2^n}-1\implies v_p(2^{2^n+1}-1)=k$$ Now, let $$ord_p(2)=d\implies d|{2^n}+1\implies 2^{n}+1=dm\implies k=v_p(2^{2^n}+1)=v_p((2^d)^m-1)=v_p(2^d-1)+v_p(m)=v_p(2^d-1),$$where the last line follows from LTE. Now we again use LTE, note that $$d|p-1\implies p-1=dl\implies v_p(2^{p-1}-1)=v_p((2^d)^l-1)=v_p(2^d-1)+v_p(l)=v_p(2^d-1)=k.$$