$I_n=\int_0^1{\frac{x^n}{x^n+1}}$. 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.

$I_n
=\int_0^1{\frac{x^n}{x^n+1}}dx
=\int_0^1{\frac{x^n+1-1}{x^n+1}}dx
=1-\int_0^1{\frac{1}{x^n+1}}dx
=1-J_n$.
Need to show that
$J_{n+1} \ge J_n
$.
$\begin{array}\\
J_{n+1} - J_n
&=\int_0^1{\frac{1}{x^{n+1}+1}}dx-\int_0^1{\frac{1}{x^n+1}}dx\\
&=\int_0^1(\frac{1}{x^{n+1}+1}-\frac{1}{x^n+1})dx\\
&=\int_0^1\frac{x^n+1-(x^{n+1}+1)}{(x^{n+1}+1)(x^n+1)}dx\\
&=\int_0^1\frac{x^n-x^{n+1}}{(x^{n+1}+1)(x^n+1)}dx\\
&=\int_0^1\frac{x^n(1-x)}{(x^{n+1}+1)(x^n+1)}dx\\
&\gt 0\\
\end{array}
$
More directly:
$\begin{array}\\
I_{n} - I_{n+1}
&=\int_0^1{\frac{x^n}{x^n+1}}dx-\int_0^1{\frac{x^{n+1}}{x^{n+1}+1}}dx\\
&=\int_0^1(\frac{x^n}{x^n+1}-\frac{x^{n+1}}{x^{n+1}+1})dx\\
&=\int_0^1\frac{x^n(x^{n+1}+1)-x^{n+1}(x^n+1)}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^{2n+1}+x^n-x^{2n+1}-x^{n+1}}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^n-x^{n+1}}{(x^n+1)(x^{n+1}+1)}dx\\
&=\int_0^1\frac{x^n(1-x)}{(x^n+1)(x^{n+1}+1)}dx\\
&\gt 0\\
\end{array}
$
