TITLE: Proof by induction: $2^n > n^2$ for all integer $n$ greater than $4$
QUESTION [12 upvotes]: I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list:

Prove, using mathematical induction, that $2^n > n^2$ for all integer n greater than $4$

So I started:
Base case: $n = 5$ (the problem states "$n$ greater than $4$", so let's pick the first integer that matches)
$2^5 > 5^2 \implies 32 > 25$ - ok!
Now, Inductive Step:
$2^{n+1} > (n+1)^2$
now expanding
$2 * 2^n > n^2 + 2n + 1$
Well, I can't go on, I don't know how to finish this proof. Could anyone help me with this one?
Also, I know that the statement (inductive hypothesis) is true. Simply, I'll have a number to an exponent $n$ twice, next to $n$ squared and $2n$ (which could be seen as quadratic and linear functions) which "grows" much lower than the exponential one. It could be easily "proved by example" with any $n$ greater than $4$.
I want to know how to prove it algebraically , so my professor does not freak out.
thanks in advance.

REPLY [12 votes]: 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 $(\spadesuit)$.