Proving $n! < n^n$ by induction for all $n\geq 2$. I am having trouble simplifying an induction question. The question is:
Let $P(n)$ be the statement that $n! < n^n$ where $n$ is an integer greater than $1$.
My work so far:
Base case $n = 2$
$2! = 2$
$2^2 = 4$
$2 < 4$. Therefore the base case is true
Inductive Hypothesis
$P(k) = k! < k^k$ for some $k > 1$
Inductive step
$P(k+1) = (k+1)! < (k+1)^{k+1}$
$k!(k+1) < k^k(k+1)$
(so from my understanding and based on previous questions, I want to take $k^k(k+1)$ into 
$(k+1)^{k+1}$ as next inductive step.
$k^k(k+1) = k^k+1 + k^k$ using distribution. 
I'm not sure where to go from here. 
Thanks! 
 A: You don't need to have equality to continue - an inequality will do. All you have to say is that $k^k<(k+1)^k$ for $k>1$.
A: $$\frac{n^n}{n!} = \frac{n^{n-1}}{(n-1)!} > \frac{(n-1)^{n-1}}{(n-1)!} = \frac{(n-1)^{n-2}}{(n-2)!} > \dots > \frac{2^2}{2!} = 2 > 1$$
A: Suppose $P(n)$ is true for an arbitrary $n \in \Bbb N$. Then $ n1 \lt n^n  $.  Using this, 
The Inductive Step goes:
$$ (n + 1)! = (n + 1) \cdot n! \lt (n + 1) \cdot n^n \lt (n + 1) \cdot (n + 1)^n $$
A: If $n!<n^n$ for $n=k$ i.e., $k^k<k!$
$(k+1)!=(k+1)\cdot k!<(k+1)k^k$
So, it is sufficient to show $(k+1)k^k<(k+1)^{k+1}\iff\left(1+\dfrac1k\right)^k>1$
which holds true for $k=1$
Use I have to show $(1+\frac1n)^n$ is monotonically increasing sequence
A: Claim: For $n\geq 2$, let $S(n)$ denote the statement
$$
S(n) : n! < n^n.
$$
Base step ($n=2$): $S(2)$ holds because $2!=2<4=2^2$.
Inductive step: Fix some $k\geq 2$ and suppose that $S(k)$ is true, where
$$
S(k) : k! < k^k.
$$
We must show that
$$
S(k+1) : (k+1)! < (k+1)^{k+1}
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
(k+1)! &= (k+1)\cdot k!\tag{by definition}\\[0.5em]
&< (k+1)\cdot k^k\tag{by $S(k)$}\\[0.5em]
&< (k+1)\cdot(k+1)^k\tag{since $k\geq 2$}\\[0.5em]
&\leq (k+1)^{k+1},
\end{align}
we see that the right-hand side of $S(k+1)$ follows. This completes the inductive step. 
By mathematical induction, $S(n)$ is true for all $n\geq 2$. $\blacksquare$
