How to prove $n < \left(1+\frac{1}{\sqrt{n}}\right)^n$ I want to know how to prove the following inequality.
For $n = 1, 2, 3, \ldots $
$$ n < \left(1+\frac{1}{\sqrt{n}} \right)^n $$
I tried with math induction but I failed. 

We have $$\left(1+\dfrac1{\sqrt{n}}\right)^n = \sum_{k=0}^n \dbinom{n}k \dfrac1{n^{k/2}} \geq \underbrace{1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6} \dfrac1{n^{3/2}} > n}_{\text{For $n \geq 3$}}$$
Check for $n=1,2$ manually.

EDIT If you do not want to do the dirty calculus of showing $$1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6} \dfrac1{n^{3/2}} > n$$ by showing that the appropriate function is increasing, here is a more logical way. We have
$$\underbrace{1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6n^{3/2}} > \dfrac{n+1}2 + \dfrac{n^{3/2}}{24}}_{\text{For }n \geq 3}$$
The above inequality comes from the fact that $\sqrt{n} > 0$, $n-1 \geq \dfrac{n}2$ and $n-2\geq \dfrac{n}2$ for $n\geq 3$. We have $\dfrac{n^{3/2}}{24} \geq \dfrac{n}2$ for $n \geq 144$. Put all this together to obtain the result for $n \geq 144$. Write a code to check if it is true for $n=1$ to $n=143$.
