Prove that $n^2 > n+1 \quad\forall n \geq 2$ using mathematical induction 
Prove $n^2 > n+1$ for $ n \geq 2$ using mathematical induction

So I attempted to prove this, but I'm not sure if this is a valid proof.
Base case, $n = 2$
$$
2^2 > 2+ 1 
$$
$n = k + 1$, induction on $n$
$$
(k+1)^2 > k + 2
$$
$k^2 + 2k + 1 > k + 2$ and by assumption, we know that $k^2 > k + 1$. so we can now write:
\begin{gather}
k^2 + 2k + 1 > 2k + 2\\
k^2 + 1 > 2\\
\end{gather}
$k^2 -1 > 0$ and since $n$ has to be $\geq 2$, the left hand side of the inequality will always be positive, making the statement a tautology. 
Does this prove the original statement? If not, how do you do it?  

We have
$$\begin{align}
k^2&\ge k+1\tag{1}\\
k^2+2k&\ge 3k+1\tag{2}\\
k^2+2k+1&\ge 3k+2\tag{3}\\
k^2+2k+1&\ge 3k+2\ge k+2\tag{4}\\
(k+1)^2&\ge (k+1)+1\tag{5}\\
\end{align}$$
