In preparation for the next sem. I am working on an old problem set, ahead of class...
Proof by induction.... This should be true for all $n \geqslant 3$
$$(n+1)! \leqslant n^n$$
I am quite new to doing proofs, and have never done a proof by induction
So I have to basically show that $$(n+2)! \leqslant (n+1)^{(n+1)}$$ By assuming that the first equation is true.
There is a comment saying that we should use the inequality of the arithmetic and geometric mean.
But no matter how hard I try i come nowhere close to either of the two expressions.
Many thanks
Base case: $n = 3$. Indeed, $4! = 24 \le 3^3 = 27$.
Suppose it is true for $k \ge 3$. Then we find:
$$(k + 2)! = (k + 2)(k + 1)! \le (k + 2) k^k = (k + 2) \cdot k \cdot k^{k - 1} \le (k + 1)^2 k^{k - 1} \le (k + 1)^2 (k + 1)^{k - 1} = (k + 1)^{k + 1}$$