Inductive Proof that $k!Call $P(k): k!<k^k$, for $k\geq 2$
Test it out with 2, and it's true ($2<4$). 
Assume that $P(k)$ is true for some $k\geq2$.
Then show that $P(k+1)$ is true. 
$P(k+1): (k+1)!<(k+1)^{k+1}$
Ministep: $(k+1)!=k!(k+1)$
Ministep: $(k+1)^{k+1}=(k+1)^{k}(k+1)$  
$k!(k+1)<(k+1)^{k}(k+1)$
Can I just pull out that (k+1) and call it a day? That will make $k!<(k+1)^{k}$. 
We're already assuming that $k!<k^k$, for some integer $k\geq 2$, and I don't think I need to prove that $k^{k}<(k+1)^{k}$. I mean, I guess I could go prove that as well, but let's assume for the moment that I don't have to. Since we're assuming the original $P(k)$ is true, I should be able to write $k!<k^{k}<(k+1)^{k}$ right? That would prove the $k+1$ case. Is this the correct way to go about things? It seems too easy, but makes sense.
Thanks for confirmation and/or pointing out mistakes!

You are right, But I like to think about it in another way.
Assume $P(k)$ is true for some $k \geq 2$ which means that $$k! < k^k$$
Now multiply each side of the inequality by $k+1$
We get that $k!(k+1) < k^k(k+1)$.
Now we know that $k!(k+1) = (k+1)! = $ LHS of the inequality
and we also know that  $k^k(k+1) < (k+1)^k(k+1) = (k+1)^{k+1}  $
And so you have $$(k+1)! < k^k(k+1) < (k+1)^k(k+1) = (k+1)^{k+1}$$ 
It's kind of easier to read through this 
