TITLE: Proof that $n^2 < 2^n$
QUESTION [18 upvotes]: 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?

REPLY [5 votes]: Assuming that $n\geq 5$, and $k^2<2^k$, $$2^{k+1}=2\cdot2^k>2k^2=k^2+k^2=k^2+2k+k(k-2)$$ Since the starting integer is $5$ so $k(k-2)$ is at least $15>1$ and therefore $$2^{k+1}>k^2+2k+1=(k+1)^2.$$