1

Basis Case:

$2! = 2\times1 = 2$

$2^2 = 4>2$

Inductive Hypothesis:

$k!<k^k$

Induction Step:

$k!<k^k$

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

$(k+1)! < k^{k+1} + k^k$

I'm confused on where to go from here, or if what I was doing was even the right way to go. I need advice on using induction for this question and on inequalities in general.

Michael Hardy
  • 1
Dunka
  • 2,827

3 Answers3

5

$$ \frac{n!}{n^n} = \frac{1 \cdot 2\cdots n}{n \cdot n \cdots n} = \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n}{n} < 1 $$ since all factors $< 1$.

Induction step:

$(n+1)! = n! (n+1) <n^n(n+1) < (n+1)^n(n+1) = (n+1)^{n+1}$

Jihad
  • 3,334
  • This is a nice proof, but do you know how to do it by induction(The problem is from the section on induction so we're expected to use induction) – Dunka Dec 01 '14 at 19:17
  • Coolest proof !!! +1 –  Jan 16 '17 at 10:56
3

$(k+1)! = k!(k+1) < k^k(k+1) <(k+1)^k(k+1) = (k+1)^{k+1}$

Michael Biro
  • 13,847
1

Easy proof by induction:

$k=2$ is true since $2<2^2=4$. Suppose the result is true for $n-1$ i.e $(n-1)!<(n-1)^{(n-1)}$. Then multiplying both sides by $n$ gives:

$n!=n(n-1)!<n(n-1)^{(n-1)}<n^{n}$

Hesky Cee
  • 452