Proof of $n^{1/n} - 1 \le \sqrt{\frac 2n}$ by induction using binomial formula Using $$(a+b)^{n} =  \sum_{i=0}^n {n \choose k} a^{n-k} b^{K}$$
prove that 
$$n^{1/n} - 1 \le \sqrt{\frac 2n}$$ for n= 2,3,4.... 
I know the first step is to set $$ n^{\frac 1n} = 1 + x $$ for some x>0 and then raise both sides to the n, but I'm lost after that.
 A: If $x=n^{1/n}-1$, then $n=(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots\ge 1+\frac{n(n-1)}{2} x^2$ for $n\ge 2$.
Then $n-1\ge \frac{n(n-1)}{2} x^2\implies 1\ge \frac{n}{2}x^2\implies x^2\le\frac{2}{n}\implies x\le\sqrt{\frac{2}{n}}$
A: Let's look at $(1+\sqrt\frac{2}{n})^n=1+n\sqrt\frac{2}{n}+\frac{n(n-1)}{2}\frac{2}{n}+\text{more positive terms}= 1+(n-1)+n\sqrt\frac{2}{n}+...\geq n$, now take the $n^{th}$ root of both sides and you get the desired results.
Two remarks: This implicitly assumes $n>2$, but you can simply check for $n<2$. I did not use induction, are you required to use induction or was it your guess that this is the way to approach this problem?
A: Here is a better result,
gotten by elementary means.
By Bernoulli's inequality,
$(1+n^{-1+1/k})^n
\ge 1+n^{1/k}
> n^{1/k}
$.
Raising to the
$k/n$ power,
$n^{1/n}
< (1+n^{-1+1/k})^k
$.
If $k = 3$,
this bound is
$(1+n^{-2/3})^3
=1+3n^{-2/3}+3n^{-4/3}+n^{-2}
<1+7n^{-2/3}
$.
Extending this,
I have shown that
$n^{1/n} < 1+2kn^{–1+1/k}$
 for 
$n>k^{k/(k-1)}$.
