$I_n=\int_0^1{\frac{x^n}{x^n+1}}$. Prove $I_{n+1} \le I_n$ for any $n \in \mathbb N$ $$I_n=\int_0^1{\frac{x^n}{x^n+1}}$$
Prove $\lim_{n\to\infty}{I_n} = 0$
Here is what I tried.
First, I rewrite $I_n$.
$$I_n=\int_0^1{1-\frac{1}{x^n+1}}=1 - \int_0^1{\frac{1}{x^n+1}}$$ 
Now the limit becomes:
$$L=1-\lim_{n\to\infty}\int_0^1{\frac{1}{x^n+1}}$$
Next I try to solve the limit of the integral using Squeeze Theorem, with no success.
Using the fact $0\le x \le 1$ I get to the following double inequality:
$$\int_0^1\frac{1}{x^{n-1}+1} \le \int_0^1\frac{1}{x^{n}+1} \le 1 $$
$$I_{n-1} \le I_n \le 1$$ 
I don't know what to do next, I would greatly appreciate some help.

For any $\epsilon \gt 0$, $\int_0^1\frac{1}{x^n+1}dx\gt K_n(\epsilon)=\int_0^{1-\epsilon}\frac{1}{x^n+1}dx$. 
For every $\delta \gt 0$, there exists an $N$, so that for all $n\gt N$, $x^n\lt \delta$.
In this case, $K_n(\epsilon)\gt\frac{1-\epsilon}{1+\delta}$.  Since  $\delta$ is arbitrarily small, $\lim_{n\to \infty}K_n(\epsilon)\ge 1-\epsilon$ so $\lim_{n\to \infty}\int_0^1\frac{1}{x^n+1}dx \ge 1-\epsilon$.  Because $\epsilon $ is arbitrary,  $\lim_{n\to\infty}\int_0^1\frac{1}{x^n+1}dx=1$ and $I_n\to 0$.
