# Proof by induction using the binomial theorem

Prove by induction that: $$\quad n!<(\frac{(n+1)}{2})^n$$ for $$n>1$$

Hint: use the binomial theorem

for $$\ n=2$$: $$\quad 2!<(\frac{2+1}{2})^2=2.25 \;$$, which is true. I tried: $$(k+1)!<(\frac{k+2}{2})^{k+1}\\ (k)!(k+1)<(\frac{k+2}{2})^k(\frac{k+2}{2})$$

And i'm basically stuck, i don't have any idea on how to continue. What does the hint mean? How can i use the binomial theorem in here? Any help would be appreciated!

To show is $$(n+1)! < \left(\frac{n+2}{2}\right)^{n+1}$$ under the assumption that $$n! < \left(\frac{n+1}{2}\right)^{n}$$ - the induction hypothesis (IH) - is true.
$$(n+1)! = (n+1)n! \stackrel{IH}{<}(n+1)\left(\frac{n+1}{2}\right)^{n}$$
So, it remains to show that $$(n+1)\left(\frac{n+1}{2}\right)^{n} \leq \left(\frac{n+2}{2}\right)^{n+1}$$ $$\Leftrightarrow 2 \left(\frac{n+1}{2}\right)^{n+1} \leq \left(\frac{n+2}{2}\right)^{n+1}$$ $$\Leftrightarrow 2 \leq \left(1+\frac{1}{n+1}\right)^{n+1}$$ which is true because of the binomial theorem since for $$n>1$$ you have $$\left(1+\frac{1}{n+1}\right)^{n+1} = \sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{(n+1)^k}\geq 1 + \frac{n+1}{n+1} = 2$$ Done.