showing an iequality without using a calculator During solving a problem, I faced the inequality $$\frac{37^{37}}{36^{36}}&lt;100$$. 
Is it useful to consider  the function $f(x)=x^x$ or something like that? Or is it obtained by algebraic manipulations? The calculator says that the two sides differ by less than $0.8$.
Please give me a hint to prove the inequality. Thanks. 

Edit. If we take logarithms of both sides, we may consider the function $f(x)=x\log x$ and apply the mean value theorem on the interval $[36,37]$. However, this does not help as we dont know the $c$ obtained by the theorem satisfies $f'(c)&lt;2$. Any other ideas?
From last "Can you answer these questions?"
$$\frac{37^{37}}{36^{36}}&lt;100\iff \ 0.37\left(\frac{37}{36}\right)^{36}\lt1\iff0.37\left(\frac{1}{36}+1\right)^{36}\lt1\\\left(\frac{1}{36}+1\right)^{36}=(a^4+4a^3+6a^2+4a+1)^9$$
$$a=\frac{1}{36}=0.0277777777....\lt0.028$$

$$\begin{cases}a\lt0.028\\a^2\lt0.000784\\a^3\lt0.000021952\\a^4\lt0.000000614656\end{cases}$$
$$\begin{cases}4a\lt0.112\\6a^2\lt0.004704\\4a^3\lt0.000087808\\a^4\lt0.000000614656\end{cases}$$ It follows $$(a^4+4a^3+6a^2+4a+1)^9=(1.116795422656)^9\lt (1.1168)^9=2.702580553...\lt2.702581$$Thus $$0.37\left(\frac{1}{36}+1\right)^{36}\lt0.37\times2.702581=0.99995497\lt1$$
I suggest rewriting ${37}^{37}$ as ${(36 + 1)}^{37}$, expand using the binomial theorem, then divide each successive term by ${36}^{36}$.  You should see the terms drop off in a suggestive pattern.