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...

$(1+x)^n \ge (n^2/4) x^2 $, $n \ge 1$, $x \ge 0$.
Proof: 
$(1+x)^n=$
$1+nx + (n(n-1)/2!)x^2+...\gt (n(n-1)/2)x^2 \ge (n^2/4)x^2,$
since we have 
$(n)((n-1)/2) \ge n( n/4)$,  for $n\ge 2.$
Let $x=2/√n$, in
$(1+x)^n > (n^2/4)x^2$;
$1+2/√n \gt \sqrt[n]{n}$, $n \ge 1$.
Finally 
$(1+2/√n)^2 >(1+2/√n) \gt \sqrt[n]{n}$, $n \ge 1$.
