Find the value of $a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$ 
Given that the sequence $\left\{a_n\right\}$ satisfies $a_0 \ne 0,1$ and $$a_{n+1}=1-a_n(1-a_n)$$ $$a_1=1-a_0$$
  Find the value of $$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$$

We actually get the terms of the sequence as:
$$a_1=1-a_0$$
$$a_2=1-a_1a_0$$
$$a_3=1-a_2a_1a_0$$
$$\vdots$$
$$a_{n+1}=1-a_na_{n-1}a_{n-2}\cdots a_0$$
Any way from here?
 A: For $n\ge 1$, we have $a_{n+1}-1=a_n(a_n-1)$.  It follows that if $a_n\notin\{0,1\}$, then $a_{n+1}\notin\{0,1\}$, but since $a_0\notin\{0,1\}$, we see that $a_n\notin\{0,1\}$ for all $n$. Now $$\frac1{a_{n+1}-1}=\frac{1}{a_n(a_n-1)}=\frac{1}{a_n-1}-\frac{1}{a_n}$$
for all $n\ge 1$. That is
$$\frac1{a_n}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$$
if $n\ge 1$, so
$$\frac1{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}=\frac1{a_1-1}-\frac{1}{a_{n+1}-1}$$
Since $a_1=1-a_0$, we get $a_1-1=-a_0$.  Thus $\frac{1}{a_1-1}=-\frac1{a_0}$ and
$$\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}=-\frac{1}{a_{n+1}-1}.$$
From your calculation, $a_{n+1}-1=-a_0a_1a_2\cdots a_n$, so
$$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}\right)=\frac{a_{n+1}-1}{a_{n+1}-1}=1.$$
A: As we get,
$$a_1=1-a_0\\
a_2=1-a_0a_1\\
\begin{align}\\
\therefore \frac{1}{a_0}+\frac{1}{a_1} &=\frac{1}{a_0}+\frac{1}{1-a_0}\\
&=\frac{1-a_0+a_0}{a_0(1-a_0)}\\
&=\frac{1}{a_0(1-a_0)}\\
&=\frac{1}{a_0a_1}\\
\end{align}$$
$$\begin{align}\\
\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2} &=\frac{1}{a_0a_1}+\frac{1}{a_2}\\
&=\frac{1}{a_0a_1}+\frac{1}{1-a_0a_1}\\
&=\frac{1-a_0a_1+a_0a_1}{a_0a_1(1-a_0a_1)}\\
&=\frac{1}{a_0a_1a_2}\\
\end{align}$$
$$.\\
.\\
.\\
.$$
$$\begin{align}\\
\frac{1}{a_0}+\frac{1}{a_1}+....\frac{1}{a_{n-1}}+\frac{1}{a_n}
&={\begin{aligned}\\
\frac{1}{a_0...a_{n-2}a_{n-1}}+&\frac{1}{1-a_0....a_{n-2}a_{n-1}}\\
\end{aligned}}\\
&=\frac{1-a_0....a_{n-2}a_{n-1}+a_0....a_{n-2}a_{n-1}}{a_0....a_{n-2}a_{n-1}(1-a_0....a_{n-2}a_{n-1})}\\
&=\frac{1}{a_0....a_{n-2}a_{n-1}a_n}\\
\end{align}$$
$$\begin{aligned}\\
\therefore a_0a_1....a_{n-1}a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+....\frac{1}{a_n}\right)
&=1\\
\end{aligned}$$
