Suppose I wish to prove that $\forall n\in\mathbb{N}\ge 5,$ the statement $\mathscr{P}(n): n^2<2^n$ is true.
Proof:
Proceeding by induction on $n$, we note that $\mathscr{P}(5):(5)^2=25<2^5=32 \implies \mathscr{P}(5) $ holds. Now we induct on $n$ and assume the statement $\mathscr{P}(n)$ holds for some $n=k>5$, that is, we assume $\mathscr{P}(k): k^2<2^{k}$ holds for some $k\in\mathbb{N}>5$. We have: \begin{align} k^2<2^k\\\ 2k^2<2^{k+1}\\\ (k^2)+k^2<2^{k+1}\\\ (2k+1)+k^2<(k^2)+k^2<2^{k+1}\\\ \end{align}
The last inequality $2k^2>k^2+2k+1$ follows from the fact that $k^2-2k-1=(k-1)^2-2>0$ for $k>5>1\pm\sqrt{2}.$ Continuing, we have that: \begin{align} k^2+2k+1<2k^2<2^{k+1}\\\ (k+1)^2<2k^2<2^{k+1}\\\ \implies (k+1)^2<2^{k+1}\space\text{as desired}\space \end{align}
Hence we have shown that the truth of $\mathscr{P}(k) \implies\space\text{the truth of}\space\mathscr{P}(k+1)\space\text{for any}\space k\in\mathbb{N}>5$. We have shown by the Principle of Mathematical Induction that $n^2<2^n$ for all $n\ge5$. $\square$
My concern with my proof is using the auxiliary notion that $k^2>2k+1$ in the induction step. Is this proof valid? Thanks a lot in advance!
This is perfectly valid, nowhere does it say that you need to use only the inductive hypothesis and nothing else. You even proved the 'auxiliary notion' to boot.