<s>[INST] 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?
 [/INST] As we get,
$$a_1=1-a_0\\
a_2=1-a_0a_1\\
\begin{align}\\
\therefore \frac{1}{a_0}+\frac{1}{a_1} &=\frac{1}{a_0}+\frac{1}{1-a_0}\\
&=\frac{1-a_0+a_0}{a_0(1-a_0)}\\
&=\frac{1}{a_0(1-a_0)}\\
&=\frac{1}{a_0a_1}\\
\end{align}$$
$$\begin{align}\\
\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2} &=\frac{1}{a_0a_1}+\frac{1}{a_2}\\
&=\frac{1}{a_0a_1}+\frac{1}{1-a_0a_1}\\
&=\frac{1-a_0a_1+a_0a_1}{a_0a_1(1-a_0a_1)}\\
&=\frac{1}{a_0a_1a_2}\\
\end{align}$$
$$.\\
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$$\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}$$
</s>