Inductive Proof that $k!<k^k$, for $k\geq 2$.

Solution 1:

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 $n\geq 2$. $\blacksquare$