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?
Hint only: For $n \geq 3$ you have $n^2 &gt; 2n + 1$ (this should not be hard to see) so if $n^2 &lt; 2^n$ then consider
$$
2^{n+1} = 2\cdot2^n &gt; 2n^2 &gt; n^2 + 2n+1 = (n+1)^2.
$$
Now this means that the induction step "works" when ever $n\geq 3$. However to start the induction you need something greater than three. By trial an error you can find the smallest $n(\geq0)$ such that $n^2 &lt; 2^n$.
As a fun fact another way to see that this inequality is true eventually is to compare the sum of the numbers in the $n-1$ 'st row of pascal's triangle to the $n$'th triangular number, which appears in the $n+1$ 'st row and occurs along one of the diagonal's of pascal's triangle.
See the link for more details, http://mathforum.org/workshops/usi/pascal/pascal_triangular.html.  The fact presented on that page plus the fact the sum of the row of pascal's triangle is a power of two can be used to prove the desired inequality.

This amounts to proving that $\frac{n(n+1)}2 \lt 2^\left(n-1\right)$ which is stronger than the original inequality, since $n^2 \lt n(n+1)$ for $n \gt 0$ .