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!
 A: 
I should be able to write $k!<k^{k}<(k+1)^{k}$ right? 

Sort of. The key part, which it sounds like you called "too easy," is realizing that you can use $k^k<(k+1)^k$ in the context of the proof; that is, this realization makes it possible to easily get from the left-hand side of what you need to prove to the right-hand side. Nonetheless, I would encourage you to take a look at this post on how to write a clear induction proof. Your proof is not as clear as it could be with the sidebar "ministeps," the fact that you never once mentioned where you used the inductive hypothesis, etc. If you were my student, then I would probably give you 8/10. My point in answering is to provide what I think, in my humble opinion, is a nice way of writing up the proof. I hope it helps. 

Claim: For $n\geq 2, n\in\mathbb{Z}, n!<n^n$.
Proof. For $n\geq 2, n\in\mathbb{Z}$, let $P(n)$ denote the statement
$$
P(n) : n!<n^n.
$$
Base case ($n=2$): $P(2)$ says that $2!=2<4=2^2$, and this is true. 
Inductive step $P(k)\to P(k+1)$: Fix some $k\geq 2$, and assume that 
$$
P(k) : \color{blue}{k!<k^k}
$$
holds. To be shown is that $P(k+1)$ follows where
$$
P(k+1) : \color{green}{(k+1)!<(k+1)^{k+1}}.
$$
Starting with the left-hand side of $P(k+1)$,
\begin{align}
\color{green}{(k+1)!} &= (k+1)\cdot \color{blue}{k!}\tag{by definition}\\[0.5em]
&< (k+1)\cdot\color{blue}{k^k}\tag{by $P(k)$, the ind. hyp.}\\[0.5em]
&< (k+1)(k+1)^k\tag{since $k<k+1$}\\[0.5em]
&= \color{green}{(k+1)^{k+1}},\tag{by definition}
\end{align}
we end up at the right-hand side of $P(k+1)$, completing the inductive step. 
By mathematical induction, the statement $P(n)$ is true for all integers $n$ where $n\geq 2$. $\blacksquare$
A: 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 
A: Both expressions are a product of $k$ factors. It's easy to see that in $k^k$, each of the factors is larger than the factors from $k!$.
I get that you are looking for an inductive proof, but any proof is trivially inductive. 
Assume $(k-1)!<(k-1)^{k-1}$. Now I need that to imply $k!<k^k$. Since $k!<k^k$, as explained above, I've done the induction step. It doesn't matter that you don't see the logical connection. It's still formally and logically true that $P(k-1)\implies P(k)$, since $P(k)$'s truth can be established independently, quickly.
