Proof that $n^2 < 2^n$ How do I prove the following statement by induction?
$$n^2 \lt 2^n$$
$P(n)$ is the statement $n^2 \lt 2^n$
Claim: For all $n \gt k$, where $k$ is any integer, $P(n)$
(since $k$ is any integer, I assume I have to prove this for positive and negative integers)
So let base case be $P(1)$, and I have to prove $P(n)$ for $n \ge 1$ and $n \lt 1$
$P(1)$ is $1^2 \lt 2^1$ which is clearly true.
Induction hypothesis: $k^2 \lt 2^k$
Inductive step $(k+1)^2 \lt 2^{(k+1)}$
$(k+1)^2 = k^2 + 2k + 1$
$k^2 + 2k + 1 \lt 2^k + 2k + 1$ by inductive hypothesis
Not sure how to proceed. Is my previous intuition that I have to prove this for $n \ge 1$ and $n \lt 1$ correct?

Your statement is only true for $n=0$, $n=1$ and $n \geq 5$.
Than start with  $n=5$, this gives $5^2=25 < 2^5 = 32$ 
So Assume it is true for $n$. Than we know that 
$$2^{k+1}= 2 \cdot 2^k > 2 k^2>k^2+2k+1=(k+1)^2$$
(as $n^2>2n+1$ for all $n\geq 3$ and as we started with $n=5$ this is always true)
