Proving $\sqrt[n]{n} < 2-\frac{1}{n}$ inductively 
I have already done the first part of the problem. For the second part they give me the following hint: Use (1) with $a =\frac{n-1}{n}$, then take $n$th roots.
I'm actually quite confused. Rewriting the expression I have that $2-\frac{1}{n}=1+\frac{n-1}{n}=1+a$. Thus $\sqrt[n]{n} < 2-\frac{1}{n} = \sqrt[n]{n} < 1+a$, so $n<(1+a)^n$. From now on I don't know how to continue with the exercise. I have made several attempts starting from the last expression but it is not clear to me

Using I) you get
$$\left(1+\frac{n-1}{n}\right)^n>1+n\frac{n-1}{n}\to \left(\frac{2n-1}{n}\right)^n>n$$
$$\frac{2n-1}{n}>n^{1/n}\to 2-\frac1n>n^{1/n}$$
