Suppose $n$ is equal or bigger than $3$. It's obviously true for $n=3$ that $n!<n^{n-1}$. To show more generally that
$$k! < k^{k-1} \text{ for some } k,$$
is it as simple as saying
$$(k+1)! = (k+1)k! \implies (k+1)!\lt k^k\:\:?$$
I'm new to induction so I am not sure I got it right. Thanks.
Showing that it's true for $3$ $$3^2\gt3!$$ Assume that it's true for $k$ $$k^{k-1}\gt k!\tag{1}$$ we can see clearly that $$(k+1)^{k-1}\gt k^{k-1}$$ $$\implies (k+1)^{k-1}\gt k!$$ Multiplying Both side by $(k+1)$ $$(k+1)(k+1)^{k-1}\gt (k+1)k!$$
$$(k+1)^{k}\gt (k+1)!$$ if it's true for $k$ then it's true for $k+1$ and then by Induction Hypothesis....