Prove using mathematical induction that $n^2 > n+1$ for all $n \ge 2$

Question

I have proved for the initial case $P(2)$ that this is true, but I'm stuck at substituting in $n=k+1$, $(k+1)^2 > (k+1)+1$ = $k^2 + 2k + 1 > k+2$, where do I go from here or have I made a mistake?

Thanks guys :)

Answer 1

Assume that $k^2>k+1$. Then

$(k+1)^2=k^2+2k+1>k+1+2k+1=3k+2>k+2,$ because $k>0$.