Simple Proof by Induction: Question About Formulation of the Inductive Step. Prove $n^2 &gt; n+1, n\geq2$:

Basis: n = 2

$n^2 &gt; n+1$

$2^2 &gt; 2+1$

$4&gt;3$

Inductive Hypothesis: Assume P(k) is true, i.e.,

$k^2 &gt; k+1$

Inductive Step: Show P(k+1) is true, i.e.,

$(k+1)^2 &gt; (k+1) + 1$

$k^2 + 2k + 1 &gt; (k+1) + 1$

$k^2 + 2k + 1 &gt; k + 2$

$k^2 + 2k &gt; k+1$

$k(k+2) &gt; k + 1$

I know that I need to try and get the expression to read $k^2 &gt; k+1$. The steps above are the closest I could get. Can some explain if my format for inductive proofs is correct and point me in the right direction to completing the proof?
Format-wise, I think what you're doing is perfectly fine. I think some more wording would be appropriate though: for instance, the way you have your inductive step written, it almost seems like you're assuming the inequality holds at every step along the way. Usually, I would start at the left-hand side and do manipulations to verify the right-hand side does hold (or the reverse). Still, don't be afraid to use words: being able to explain what you're doing and why is very important in proof-writing!



So, on the premise $k^2 &gt; k+1$, you wish you show $(k+1)^2 &gt; k+2$. We see that

$$(k+1)^2 = k^2 + 2k + 1$$

Invoking the induction hypothesis ($k^2 &gt; k+1$), we see that

$$k^2 + 2k + 1 &gt; k+1 + 2k + 1 = 3k + 2 &gt; k+2$$

(for which the inequality holds since $k$ is assumed positive). Thus, $(k+1)^2 &gt; k+2$ as desired.
$$(k+1)^2=k^2+2k+1&gt;k+1+2k+1&gt;(k+1)+1.$$