Q:

Proving $f(n + 1) > f(n)$ and is f injective?

If I have a function $f:\mathbb N \to \mathbb N$ defined for every $n \in \mathbb N$  by:
$$f(n) = (n+1)!-1$$
How would I prove that $f(n+1) > f(n)$ for every $n\in\mathbb{N}$? 
Would it be sufficient to show that $f(n+1) > f(n)$ is true for the lowest element that n can be?
And how would I prove that f is indeed injective?

A:

Since
$$(n+2)!=(n+2)(n+1)!\ge2(n+1)!>(n+1)!,\quad \forall n\ge0$$
then we see that $f(n+1)>f(n)$ and then $f$ is injective since by definition
$$n\ne m\implies f(n)\ne f(m)$$

