Induction inequality check: $n!<n^n$

Question

check my proof, I feel like I made a mistake :)

so I'm looking to prove that when $p(n)$ is $n!<n^n$, $p(n)$ is true for all $n>1$.

Base Case $$ p(2) \iff 2!<2^2 \iff 2<4 $$

Assume p(k) is true $$k!<k^k$$

Prove p(k+1)

$$(k+1)!<(k+1)^{(k+1)}$$ $$(k+1)(k)!<(k+1)(k+1)^k$$ $$(k+1)(k)!<(k+1)(k^k+1)$$ This part above. Can I assume that $1^k$ is always $1$ given any $k$ such that $k>1$? $$(k+1)(k)!<(k+1)(k^k+1)$$ $$(k)!<k^k+1$$ Then above, can I factor out a k+1 from both sides? $$(k)!<k^k+1$$

Is this a completed proof? Would my ending statement be something like "since we assumed $p(k)$ and $p(k+1)$ is still true given $p(k)$, and since $p(k+1)$ is a higher degree than $p(k)$...." (not sure what really to say here)?

trying again

prove p(k+1)

to start, im now looking to multiply the sides by (k+1)? not replace?

so

$$k!<k^k$$ $$(k+1)k!<(k+1)k^k$$ $$(k+1)!<(k^[k+1]+k^k$$

Answer 1

Finishing up the induction proof, \begin{align} (k+1)!&=(k+1)k! \\ &<(k+1)k^k & \text{by hypothesis, } k! < k^k\\ &<(k+1)(k+1)^k & k < k+1 \text{ for all }k>1\\ &=(k+1)^{k+1} \end{align}

Answer 2

As an alternative solution, we could use the AM-GM inequality:

$$\frac{1 + 2 + 3 + \dots + n}{n} \ge \sqrt[n]{n!}$$ $$\frac{n(n+1)}{2n}\ge \sqrt[n]{n!}$$ $$\frac{n+1}{2}\ge\sqrt[n]{n!}$$

Since $n> \frac{n+1}{2}$ for $n>1$, we have

$$n>\sqrt[n]{n!}$$

And finally,

$$n^n > n!$$