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...

Beware: overkill incoming. We may use the AM-GM inequality for producing tight bounds for $\sqrt[n]{n}$. In particular, by noticing that
$$ n = 1\cdot \frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n}{n-1} $$
we have
$$ \sqrt[n]{n}=\text{GM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)<\text{AM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right) $$
and 
$$ \sqrt[n]{n} < 1+ \frac{H_{n-1}}{n},\qquad \left(1+\frac{1}{\sqrt{n}}\right)^2=1+\frac{2\sqrt{n}+1}{n}, $$
so it is enough to show that $H_{n-1}\leq 2\sqrt{n}+1$, at least from some point on. On the other hand the Cauchy-Schwarz inequality ensures
$$ H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}\leq\sqrt{\sum_{k=1}^{n-1}1\sum_{k=1}^{n-1}\frac{1}{k^2}} < \sqrt{n\zeta(2)} $$
and we are done.
