<s>[INST] Showing that $\sqrt[105]{105}>\sqrt[106]{106}$ How can one prove that 
$$\sqrt[105]{105}>\sqrt[106]{106}\text{ ?}$$
Induction on the statement $$\sqrt[n]{n}>\sqrt[n+1]{n+1} \text{ for } n \in \mathbb{N}| n>2$$ would yield $\sqrt[3]{3}>\sqrt[4]{4}$ at the base step $n=3$, which we cannot assume. 
So, based on the properties of powers and square roots alone, can we prove the first statement? 
EDIT: No calculus, no functions, not even logs.
It's more of a riddle that anything else.
 [/INST] All you need here is the binomial theorem for positive integer exponents:
$$(1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$$
Fix $n \ge 3$, and put $x = \frac{1}{n}$:
$$\left(1+\frac{1}{n}\right)^n = \sum_{r=0}^n \binom{n}{r}n^{-r}$$
Now, $\binom{n}{r} \le n^r$ for all $r=0,\ldots,n$ (easily proven by induction; or just look at the expression $\binom{n}{r} = \frac{n}{r}\frac{n-1}{r-1}\cdots\frac{n-r+1}{1}$). So
$$\left(1+\frac{1}{n}\right)^n \le \sum_{r=0}^n n^rn^{-r} = n+1$$
This is not quite good enough for our purposes, but we can improve the estimate by considering the last two terms separately:
$$\begin{align}
\left(1+\frac{1}{n}\right)^n &\le \sum_{r=0}^{n-2} n^rn^{-r} + \binom{n}{n-1}n^{-(n-1)}+\binom{n}{n}n^{-n} \\
&= n - 1 + n^{-(n-2)}+n^{-n} \\
&< n - 1 + 2n^{-(n-2)} \\
&< n
\end{align}$$
because $n^{n-2} > 2$ if $n \ge 3$, so $n^{-(n-2)} < \frac12$.
Therefore $n > (1+\frac{1}{n})^n$ for all $n \ge 3$. Multiplying both sides by $n^n$ gives $n^{n+1} > (n+1)^n$. Taking the $n(n+1)$-th root gives, for all $n \ge 3$,
$$\sqrt[n]{n} > \sqrt[n+1]{n+1}$$
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