Proof by induction that $2^n - 1 > n^2$ i want to prove that $\forall n\geq 5$ 
$$2^{n}-1 > n^{2}$$
so the basis is trivial, and in the induction step (n+1), i stuck. i get :  
$(n+1)^{2} =  n^{2} + 2n + 1 < (2^{n} -1)+ 2n+1 = 2^{n} + 2n $
what i need to do from here?
tnx!

$$p(n): 2^n-1>n^2\\p(n+1):2^n-1>(n+1)^2\\
$$multiply p(n)  by 2
$$2(2^n-1>n^2)\\2^{n+1}-2>2n^2\\2^{n+1}-1>2n^2+1\\$$
no try to prove $$2n^2+1>(n+1)^2 \\2n^2+1>n^2+2n+1\\n^2-2n >0\\n(n-2)>0\\$$it is correct because n>5
