Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality: 
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...

It is equivalent to prove 
$$ n < \left(1 + \frac{1}{\sqrt{n}}\right)^{2n}.$$
By the binomial identity, the right hand side is at least
\begin{align*}
& 1 + \binom{2n}{1}\frac{1}{\sqrt{n}} + \binom{2n}{2}\frac{1}{n} \\
= & 1 + 2\sqrt{n} + 2n - 1 \\
= & 2n + 2\sqrt{n} \\
> & n.
\end{align*}
