Elementary lower bounds for $n^{1/n}$ I can show that
$n^{1/n} > 1+1/n$
for integer $n \ge 3$
by completely elementary means - 
no logs, exponentials,
or calculus.
Are there better bounds that
can be proved in an elementary way?
Here is my proof:
The bound is equivalent to
$n > \frac{(n+1)^n}{n^n}$
or
$\frac{(n+1)^n}{n^{n+1}} < 1$.
For $n=3$,
this is $\frac{4^3}{3^4} 
=\frac{64}{81}
< 1$.
The ratio of consecutive terms is
$\begin{align}
\frac{\frac{(n+2)^{n+1}}{(n+1)^{n+2}}}{\frac{(n+1)^n}{n^{n+1}}}
&=\frac{(n(n+2))^{n+1}}{(n+1)^{2n+2}}\\
&=\frac{(n^2+2n)^{n+1}}{(n^2+2n+1)^{n+1}}\\
&< 1
\end{align}
$
so the terms are decreasing
and thus less than $1$.
Of course, since $e^x > 1+x$,
$n^{1/n} = e^{\ln n/n} > 1+\ln n/n$,
but this is non-elementary.

$$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac{n!}{k!(n-k)!\cdot n^k}$$
After cancelling $n!$ against $(n-k)!$ we are left with $k$ factors $\le n$ in the numerator. Since we have $k$ factors $=n$ in the denominator, we have $\frac{n!}{k!(n-k)!\cdot n^k}\le \frac1{k!}$. This brings us almost to the series $\sum \frac 1{k!}$ for $e$, but remains elementary enough, I guess. All but the $0$th summand are $\le \frac12$, so we get the very weak $\left(1+\frac1n\right)^n\le 1+\frac n2$, hence for $n> 2$
$$\sqrt[n]n>\sqrt[n]{1+\frac n2} \ge 1+\frac1n.$$
Of course, we can easily improve this by noting $\frac1{k!}<\frac1{2^{k-1}}$ for $k\ge1$ and hence $\left(1+\frac1n\right)^n<3$.
This makes
$$ \sqrt[n]n\ge\sqrt[n]9> \left(1+\frac1n\right)^2=1+\frac2n+\frac1{n^2}\qquad \text{for }n\ge 9$$
and so on.
