Proving $2^{n+1} < n^2 + 2$ for $n\geq 0$ by induction I'm trying to prove that $2^{n+1} < n^2 + 2$ for $n \ge 0$ by use of mathematical induction, but I get to the inductive step and get lost. I don't know how to link my assumption to the proof.
 A: The question was solved in comments. I am posting a CW-answer so that it does not remain unanswered.
You can check that the inequality in the question if false even for very small numbers:
$$
\begin{array}{|c|c|c|}
\hline
  n & 2^{n+1} & n^2+2 \\\hline
  1 & 4 & 3 \\\hline
  2 & 8 & 6 \\\hline
  3 &16 &11 \\\hline
  4 &32 &18 \\\hline
\end{array}
$$
However, if you turn the inequality sign, you should be able to show that it is true by induction. Have a look at proofs of very similar inequality given in answers here: Proof that $n^2 < 2^n$ 
A: Try writing the inequality the other way: $2^{n+1}>n^2+2$. It is clear this is not true for $n=0$, but try to prove it is true for $n\geq 1$. For $n=1$, we have that $2^{n+1}=2^{2}=4>3=1^2+2=n^2+2$. Now show the inequality holds for when $n=2$: we have $2^{3}=8>6=2^2+2$, and this is true. Now try a proof by induction. I'll provide a brief sketch below.
Claim: For all $n\geq 1, 2^{n+1}>n^2+2$. Let this claim be denoted by $S(n)$. 
Proof. We checked the base cases for $n=1,2$. Now, for some fixed $k\geq2$, assume that 
$$
S(k) : 2^{k+1}>k^2+2
$$
holds. We must now use this assumption (called the inductive hypothesis) to prove that
$$
S(k+1) : 2^{k+2}>(k+1)^2+2
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
2^{k+2} &= 2(2^{k+1})\tag{exponent law}\\[0.5em]
&> 2(k^2+2)\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&= 2k^2+4\tag{simplify}\\[0.5em]
&> k^2+2k+3\tag{since $k\geq 2$}\\[0.5em]
&= (k^2+2k+1)+2\tag{rearrange}\\[0.5em]
&= (k+1)^2+2,\tag{factor}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step. 
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. 
Did all of that make sense? If a step was unclear, feel free to leave a comment. 
