Prove that $\sqrt[2012]{2012!}<\sqrt[2013]{2013!}$

Question

I need to prove that $\sqrt[2012]{2012!}<\sqrt[2013]{2013!}$

My attempt:

Let $a=\sqrt[2012]{2012!}$ and $b=\sqrt[2013]{2013!}$

Then $\displaystyle\frac{b^{2012}}{a^{2012}}=\frac{2013}{b}$

Clearly $\displaystyle b<2013$ so $\dfrac{2013}{b}>1\implies \dfrac{b^{2012}}{a^{2012}}>1\implies b>a$

I want to know if this is valid and if there is a better way of proving it.

Answer 1

For integer $n>0$, $$\{(n+1)!\}^{\frac1{n+1}}> \{n!\}^{\frac1n}\iff \{(n+1)!\}^n> \{n!\}^{n+1}$$ (Taking lcm$(n,n+1)=n(n+1)$th power in either side)

$$\iff(n+1)^n\cdot\{n!\}^n>\{n!\}^{n+1} \iff (n+1)^n>n! $$ which is true as $n+1>r$ for $1\le r\le n$

Answer 2

My proof: ($n=2012$, $n+1=2013$). $$\begin{align}((n!)^{\frac{1}{n}})^n &< ((n+1)^{\frac{1}{n+1}})^n \\ n! &< ((n+1)!)^{1-\frac{1}{n+1}} \\ n! &< (n+1) \cdot n! \cdot (n+1)!^{-\frac{1}{n+1}} \end{align}$$ Since $(n+1)!^{\frac{1}{n+1}} < n+1$, we have $(n+1)!<(n+1)^{n+1}$.