$n! \le n^n$
I did this:
$$k! \le k^k, \\ (k+1)! \le (k+1)^{k+1} \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot k \cdot (k+1) \le(k+1)^{k+1} \iff \\ k! \le (k+1)^{k}$$
Since I assumed that $k! \le k^k$ and $k+1 > k$, this has to be true.
Is this correct?
On
We are asked to prove that $n! \le n^n$ for all natural numbers $n$ by mathematical induction. To do this, we will proceed by these three steps
Inside the proposed proof, the OP is missing two of these steps. The first step that is missing from the proof is the base step. We have that $1\le 1^1=1$ so the inequality is true for $n=1$.
The next step that is missing from the proof is the induction hypothesis. We will assume that the inequality is true for $n=k$. That is, $k! \le k^k$ for all natural numbers $k$.
The final step of proof is the induction step. We need to show that the inequality is true for $n=k+1$, so we need to prove that $(k+1)! \le (k+1)^{k+1}$. In the second step of the proof made by the OP, an assumption was made
$$ (k+1)! \le (k+1)^{k+1}$$
which was then used to show that the inequality was true. I suggest starting from the left-hand side of this inequality and then deducing that the left-hand side must be less than or equal to the right-hand side. This can be accomplished by writing
\begin{align} (k+1)! &= (k+1)k! \\&\le (k+1)k^k & \text{since $k! \le k^k$ by the induction hypothesis} \\&\le (k+1)(k+1)^{k} & \text{as $k^k \le (k+1)^{k}$ for all natural numbers $k$ } \\&= (k+1)^{k+1} \end{align}
therefore, by the principle of mathematical induction, we have shown that $n! \le n^n$ for all natural numbers $n$.
You proof is conceptually correct, but the order of exposition is weird. I would write:
$$k!\le k^k\implies k!\le(k+1)^k\implies (k+1)!\le(k+1)^{k+1}.$$
The base case can be
$$1!\le 1^1.$$