how to compute this limit 
compute $I=\lim\limits_{n\to+\infty}\sqrt[n]{\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx}$

attempt: I tried to evaluate the integral
$$\begin{align}
\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx&=\int\limits_0^1x^{n+1}\left(1-x-x^2-x^3+x^5+\cdots\right)dx\\
&=\int\limits_0^1x^{n+1}-x^{n+2}-x^{n+3}-x^{n+4}+x^{n+6}+\cdots dx\\
&=\left.\frac{x^{n+2}}{n+2}-\frac{x^{n+3}}{n+3}-\frac{x^{n+4}}{n+4}-\frac{x^{n+5}}{n+5}+\frac{x^{n+6}}{n+6}+\cdots\right|_0^1\\
&=\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}-\frac{1}{n+5}+\frac{1}{n+6}+\cdots
\end{align}$$
attempt 2:
I think I can use mean value theorem for integrals

if $f$ are continuous into $[a,b]$ then exists $\xi\in[0,1]$ such that $\int\limits_a^b f(x)dx=f(\xi)(b-a)$

then using it for $[a,b]=[0,1]$ since the function inside integral are continuous, then exists $\xi\in[0,1]$ such that 
$$\begin{align}
\sqrt[n]{\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx}&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)(1-0)}\\
&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}
\end{align}$$
then making for $n\in\mathbb{N}^*$
$$\begin{align}
a_n&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}\\
a_{n+1}&=\sqrt[n+1]{\xi^{n+2}(1-\xi)\cdots(1-\xi^n)(1-\xi^{n+1})}\\
&=\sqrt[n+1]{\xi\xi^{n+1}(1-\xi)\cdots(1-\xi^n)(1-\xi^{n+1})}\\
&=\sqrt[n+1]{\xi\left[\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}\right]^n(1-\xi^{n+1})}\\
&=\sqrt[n+1]{a_n^n\xi(1-\xi^{n+1})}\\
a_{n+1}^{n+1}&=a_n^n\xi(1-\xi^{n+1})
\end{align}$$
and $\xi\in(0,1)\Rightarrow\xi^n\in(0,1)\Rightarrow 1-\xi^n\in(0,1)\Rightarrow a_n>0$ and $a_n\in(0,1)$
then
$$\begin{align}
a_{1}&=\xi^2(1-\xi)\\
a_{2}&=\sqrt{\xi^3(1-\xi)(1-\xi^2)}\\
&=\sqrt{\xi^3(1-\xi)^2(1+\xi)}\\
&=\xi(1-\xi)\sqrt{\xi(1+\xi)}
\end{align}$$
for $\xi\in(0,1)\Rightarrow \xi(1-\xi)>0$
$$\begin{align}
0<\xi&<1+\xi\\
0<\xi^2&<\xi(1+\xi)=\xi+\xi^2\\
\xi&<\sqrt{\xi(1+\xi)}\\
\xi^2(1-\xi)&<\xi(1-\xi)\sqrt{\xi(1+\xi)}\\
a_1&<a_2
\end{align}$$
then i think if i can proof that $a_{n+1}>a_n,n\in\mathbb{N}^*$ and $\xi\in(0,1)$ then the limit would be $1$

We have:
$$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$
while on the other hand, if we set:
$$ A(n,m)\triangleq\int_{0}^{1} x^n \prod_{k=1}^{m}(1-x^k)\,dx,\qquad B(n,m)=\int_{0}^{1} x^n (1-x^m)^m\,dx \tag{2}$$
we have:
$$\begin{eqnarray*} I_n = A(n+1,n) &=& A(n+1,n-1)-A(2n+1,n-1)\\&=&A(n+1,n-2)- A(2n,n-2)-A(2n+1,n-1)\end{eqnarray*} $$
and since $A(2n+2-i,n-i)\leq B(2n+2-i,n-i)\leq\frac{1}{(n+2)^3}$ it follows that:
$$ I_n = A(n+1,1)-\sum_{k=1}^{n-1}A(2n+2-k,n-k) \geq \frac{1}{(n+2)(n+3)}-\frac{n-1}{(n+2)^3}\tag{3}$$
so $I_n\geq\frac{2}{(n+2)^3}$ and this inequality, together with $(1)$, implies:
$$ I_n^{\frac{1}{n}}\xrightarrow[n\to +\infty]{}\color{red}{1}. \tag{4}$$
