I am trying to solve this excercise for induction:
Prove that for every positive integer $n > 1$, $n! < n^n$.
The first thing that I did is to prove $P(2)$: $$ P(2):\ 2 × 1 <2^2, $$ And this is true because $2 < 4$.
Assuming that $P(k)$ is true for some $k$ in general, I do not know how to prove it for $k + 1$.
Can you help me please?
On
Here is the induction step:
Assume, for some $k>1$, that $$ k!<k^k $$ Now we have $$ (k+1)! = k!\cdot (k+1)<k^k\cdot (k+1)<(k+1)^k\cdot (k+1) = (k+1)^{k+1} $$
On
Base case, $n=2$, gives $2!=2<4=2^2$, so that checks out just fine.
Now assume $$n!<n^n\tag{*}$$ for $n>1$. We have
$$(n+1)!=(n+1)n!<(n+1)n^n,$$ by assumption (*). But then $$(n+1)n^n<(n+1)(n+1)^n=(n+1)^{n+1}.$$
Since the base case holds, so do all consecutive cases, and the inequality holds in general.
You got the Base case that $1\cdot 2 <2^2$. Now we assume that the statement is true for some $n=k$, and so: $$k!<k^k$$ Is assumed true. Now we use this to prove the statement true for $n=k+1$, $$(k+1)!<(k+1)^{k+1}$$Note that $(k+1)!=(k+1)k!$ and $(k+1)^{k+1}=(k+1)(k+1)^k$. We use these to rewrite the inequality as: $$(k+1)k!<(k+1)(k+1)^k$$ The $(k+1)$ is common, so get rid of them. $$k!<(k+1)^k$$ and notice that for $k>0$ (which it is), $(k+1)>k \to (k+1)^k>k^k$ which means that $$k!<(k+1)^k$$ holds true.