Proof by induction: $2^n > n^2$ for all integer $n$ greater than $4$ [duplicate]
You proved it's true for $n=5$. Now suppose it's true for some integer $n\geq 5$. The aim is to prove it's true for $n+1$. But $$(\spadesuit)\quad2^{n+1}=2\times 2^n>2\times n^2>(n+1)^2.$$ The first inequality follows from the induction hypothesis and as for the second, we know that $(n-1)^2\geq4^2>2$, since $n\geq 5$. We can expand this inequality $(n-1)^2>2$ as follows: \begin{align*} n^2-2n+1>&\,2\\ n^2-2n-1>&\,0\\ 2n^2-2n-1>&\,n^2\\ 2n^2>&\,n^2+2n+1=(n+1)^2, \end{align*} which is the second inequality claimed in $(\spadesuit)$.