Showing that $\sqrt[105]{105}>\sqrt[106]{106}$ How can one prove that 

$$\sqrt[105]{105}&gt;\sqrt[106]{106}\text{ ?}$$
Induction on the statement $$\sqrt[n]{n}&gt;\sqrt[n+1]{n+1} \text{ for } n \in \mathbb{N}| n&gt;2$$ would yield $\sqrt[3]{3}&gt;\sqrt[4]{4}$ 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.
One way to do this is to simply note that the function $x^{1/x}$ is decreasing over a range including $105$ and $106$.

To this end, it suffices to show that $\ln[x^{1/x}] = \frac{\ln(x)}{x}$ is decreasing over such a range.

We calculate:
$$
\frac{d}{dx}\left( \frac{\ln(x)}{x} \right) = 
\frac{1 - \ln(x)}{x^2}
$$
So, this function is decreasing when $1 - \ln(x) &lt; 0 \implies x &gt; e$.  So, the function is decreasing over the range $(e,\infty)$, which suffices for our purposes.



For that base step, should you want to prove this without calculus:
$$
3^{1/3} &gt; 4^{1/4} \iff\\
3^4 &gt; 4^3 \iff \\
81 &gt; 64 \quad \checkmark
$$
Graphing $y=x^{\frac{1}{x}}$ , Y has a maximum value at x = e. 



It's all asymptotic to 1 as x approaches infinity



Updated with graph to show exact area of interest. When properly scaled, it's clear that Y is dropping like a rock as X increases. The good answers here still don't adhere to Op's request for "powers and square roots alone" and his edit of "No calculus, no functions, not even logs."
The highest rated answer? ln x from the start.