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

Your very close. Note that
$$(1+a)^n>1+na=1+n\frac{n-1}{n}=1+n-1=n$$
Can you finish it from here?
