proof with induction $2^n + 4 > n^2 + 2n$ prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof  $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here

Please note this is not true for low values of $n \in \mathbb{Z+}$ up to $4$.
Just your induction step -
If $2^{n} + 4 > n^2 + 2n$ is true
we need to prove $\, 2^{n+1} + 4 > (n + 1)^2 + 2(n+1)$
$2^{n+1} + 4 = 2 \times 2^{n} + 8 - 4 = 2 (2^{n} + 4) - 4$
$ \gt 2n^2 + 4n - 4 = n^2 + (n+1)^2 + 2(n+1) - 7 = (n+1)^2 + 2(n+1) + (n^2 - 7)$
As $(n^2 - 7) \gt 0$ for $n \geq 3$, this is true.
