How to prove (if possible) $n^2 \in o(n!)$?

Question

I'd like to prove $n^2 \in o(n!)$ for $n \in \mathbb{N}$. More specifically, this means $$ \exists c \in \mathbb{R}, n_0 \in \mathbb{N} \forall n \geq n_0: n^2 < c \cdot n! $$

My ideas

I've initially considered a proof by induction, using $c=1$ and $n_0 = 4$. In the induction step (i.e. considering $n > n_0$, I find that $$ n^2 + 2n + 1 < n! + 2n + 1 $$ However, I still have to conclusively show that $n! + 2n + 1 < (n+1) \cdot n!$. This fact seems rather obvious, however I am unable to express it.

Is there a better way to go about this task?

Answer 1

As Kanu said in the comment, this can be done directly. But to follow your approach, $n! + 2n + 1 < (n+1) \cdot n!$ is equivalent to $2n+1 < n \cdot n!$. Now since clearly $n! > 3$ and $2n+1 < 3n$ for $n \geq 3$, the latter inequality follows: $$2n+1 < 3n < n! \cdot n.$$

Answer 2

By limit definition we have

$$n^2 \in o(n!) \iff \lim_{n\to \infty} \frac{n^2}{n!}\to 0$$

and since $n!>n^3$ for $n\ge 6$ we have that $$\frac{n^2}{n!}\le \frac1n \to 0$$