Prove $2^n > n^2 + n + 1$ I did this but I'm pretty sure it's wrong
$P(5) $ is true
Assume $P(k)$ is true
for some $k \ge 5$, ie. $2^k > k^2 + k + 1$
$$2^{k+1} = (k + 1)^2 + (k + 1) + 1$$
$$2^{k+1} = k^2 + 3k + 3$$
Then:
$$2^{k+1} > (k^2 + k + 1) \cdot 2 > k^2 + 3k + 3$$
$$2^{k+1} > k^2 > k + 1$$
From the onset we know that:
$$2^{k+1} > 2^k > k^2 + k + 1$$
Therefore:
$$2^{k+1} > 2^k$$ (which is true)
Therefore, by the PMI, $P(k)$ is true for all $k \ge 5$

You started well but the end of your proof is not right and it is unclear when you write 2 inequalities.
We want to prove $2^{k+1}>(k+1)^2+(k+1)+1$ knowing already that $2^k>k^2+k+1$.
$2^{k+1}=2\times2^k>2\times(k^2+k+1)=2k^2+2k+2$ but we wanted to prove $2^{k+1}>(k+1)^2+(k+1)+1=k^2+3k+3$ so it is enough to prove that $2k^2+2k+2>k^2+3k+3$.
Which is the same as $k^2>k+1\iff k(k-1)-1>0$ which is always true for any $k\geq5$.
Then $2^{k+1}>2\times(k^2+k+1)>(k+1)^2+(k+1)+1$ and so the proof is done.
