Proof of a Basic Inequality I am new to this stack exchange and if I have any wrongdoing please let me know. My question is how to prove the following inequality:
$2^{n+1}>n^2$  assuming $n \in \mathbb{N}$
My thought is to prove this by mathematical induction.
Let $P(n)$ be the proposition
$P(1)$ is true as $4 = 2^2 > 1^2 = 1$
Assume $P(k)$ is true for some positive integer $k$
i.e. $2^{k+1}>k^2$
then $2^{(k+1)+1} = 2 \cdot 2^{k+1} > 2 \cdot k^2$ (By induction assumption)
But I get difficulty here. How can I show that $2 \cdot k^2 > (k+1)^2$ such that $P(k+1)$ is true? Thank you for your help.

$2^{n+1}>n^2$ is true for $n\in\{1,2,3\}$. Let $n\ge 4$. By Binomial Theorem:
$$2^{n+1}=(1+1)^{n+1}=\binom{n+1}{0}+\binom{n+1}{1}+\cdots +\binom{n+1}{n+1}$$
$$>2\cdot \binom{n+1}{2}=n(n+1)>n^2$$
