Question: Hello I am looking for some advice on the following,

I am wanting to show that $n^{\frac{1}{n}} \lt (1+\frac{1}{\sqrt{n}})^{2}$ for all $n \in \mathbb{N}$

and I thought I would try by induction 

The base case of n=1 is clear because $1 \lt 4$

Now I said suppose it holds that

$$n^{\frac{1}{n}} \lt (1+\frac{1}{\sqrt{n}})^{2}$$

then I must show this implies the truth of

$$(n+1)^{\frac{1}{n+1}} \lt (1+\frac{1}{\sqrt{n+1}})^{2}$$

I used Bernoulli to show that $(1+\frac{1}{\sqrt{n+1}}^{n+1} \gt \sqrt{n+1}$ for all n and then I thought if I can just use that and show that $\sqrt{n+1} \gt (n+1)^{\frac{1}{n+1}}$ then my proof would be complete.

Any advice guys? How does it look? I tried to do as much as I can on my own as I do want to learn, but I also dont want to be coming up with false proofs etc

Here is my new way of saying it,

We already showed base case,

now suppose that $$n^{1/n} \lt (1+\frac{1}{\sqrt{n}})^{2}$$

Then we want to show $$(n+1)^{1/(n+1)} \lt (1+\frac{1}{\sqrt{n+1}})^{2}$$

I then use $$(1+\frac{1}{\sqrt{n+1}})^{2} \ge 1+2\sqrt{n+1} \gt \sqrt{n+1} \gt (n+1)^{\frac{1}{n+1}}$$ to show it holds,

how does this seem?

Answer: 
Applying the AM-GM inequality:

$\sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdot 1\cdots 1\cdot \sqrt{n}\cdot \sqrt{n}}\leq \dfrac{n-2+2\sqrt{n}}{n}<1+\dfrac{2}{\sqrt{n}}<1+\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}=\left(1+\dfrac{1}{\sqrt{n}}\right)^2$
Hint: Equivalently we want to show that $n\lt \left(1+\frac{1}{\sqrt{n}}\right)^{2n}$.

For $n\ge 2$, use the first three terms of the binomial expansion of $\left(1+\frac{1}{\sqrt{n}}\right)^{2n}$.