This was a question in a maths contest, where no calculator was allowed. Also, note that only a (>,< or =) relationship is being searched for and not the value of the numbers itself.
Which is larger, $\sqrt[2015]{2015!}$ or $\sqrt[2016]{2016!}$ ?
What I've done:
My approach is to divide one number by the other and infer from the result which number is the bigger one;
WolframAlpha gives $\frac{\sqrt[2016]{2016!}}{\sqrt[2015]{2015!}}=1.0049\ldots$, so clearly $\sqrt[2016]{2016!}>\sqrt[2015]{2015!}$
Let $a=\sqrt[2016]{2016!}$ and $b=\sqrt[2015]{2015!}$
$\therefore a=\sqrt[2016]{2016!}={2016!}^{1 \over 2016}=2016^{1 \over 2016}\times2015!^{1\over 2016}=\sqrt[2016]{2016}\cdot \sqrt[2016]{2015!}$
$\therefore b=\sqrt[2015]{2015!}={2015!}^{1 \over 2015}={2015!}^{\frac{2016}{2015}\cdot\frac{1}{2016}}=\sqrt[2016]{2015!^{2016 \over 2015}}$
Hence
$$\begin{align}
\require{cancel}
\frac{a}{b}=\frac{\sqrt[2016]{2016!}}{\sqrt[2015]{2015!}}&=\frac{\sqrt[2016]{2016}\cdot \sqrt[2016]{2015!}}{\sqrt[2016]{2015!^{2016 \over 2015}}}\\
&=\sqrt[2016]{2016}\cdot \sqrt[2016]{2015!^{\frac{-1}{2015}}}= \cancelto{*}{\sqrt[2016]{\frac{2016}{2015!^{2015}}} \quad \text{which appears to be} <1}\\
=\sqrt[2016]{\frac{2016}{2015!^\frac{1}{2015}}}\\
\end{align}$$
That is $\cancelto{*}{\frac{a}{b}<1 \implies a<b}$ which is false as per the result from WA.
EDIT:
*: Correction due to error pinpointed out by Daniel Fischer.
But now I'm stuck; how do I infer what value is $\sqrt[2016]{\frac{2016}{2015!^\frac{1}{2015}}}$?
So, where did I go wrong?. How do I proceed now?
Starting with:
$$\sqrt[2015]{2015!}\mid\sqrt[2016]{2016!}$$
Raise each side to the power of $2015\cdot2016$:
$$2015!^{2016}\mid2016!^{2015}$$
Divide each side by $2015!^{2015}$:
$$2015!^{1}\mid2016^{2015}$$
Write it explicitly as:
$$\underbrace{1\cdot2\cdot\ldots\cdot2015}_{2015\text{ terms}}\mid\underbrace{2016\cdot2016\cdot\ldots\cdot2016}_{2015\text{ terms}}$$
Obviously, the RHS is larger.