Q:

Prove that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$

I am having difficulty proving the following inequality:
$$
\sqrt[n]{n} < 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n.
$$
I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa. 
I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!

A:

We shall show that $\,\Big(1+\sqrt{\frac{2}{n}}\,\Big)^n>n$. 
For $n=1$ it is obvious. 
Assume that $n\ge 2$, then according to the Binomial Formula:
\begin{align}
\left(1+\sqrt{\frac{2}{n}}\right)^n&=\binom{n}{0}+\binom{n}{1}\sqrt{\frac{2}{n}}+\binom{n}{2}\left(\sqrt{\frac{2}{n}}\right)^{\!2}+\cdots+\binom{n}{n}\left(\sqrt{\frac{2}{n}}\right)^{\!n} \\
&\ge 1+n\sqrt{\frac{2}{n}}+\frac{n(n-1)}{2}\cdot\frac{2}{n}=n+\sqrt{2n}>n.
\end{align}

