solution to a specific differential equation Does this equation have a known solution given a starting value $\phi_1$?
$$\phi_{n}\prod_{j=1}^{n-1}(1-\phi_{j}^2) + \phi_{n-1}\phi_1=0$$
Background: I am trying to derive the partial autocorrelation function of a first-order moving average process $Y_t=\epsilon_t+\theta\epsilon_{t-1}$. There, 
$$\phi_1=\frac{\theta}{1+\theta^2}$$
and the goal is to show that
$$
\phi_n=\frac{-(-\theta)^n}{1+\theta^2+\cdots+\theta^{2n}}
$$
See e.g. my partial answer here.
EDIT:
@Elaqqad, I cannot replicate that the formula is wrong for $n=4$ (or $n=5$ for that matter):

 A: There is another recurrence relation for your $\phi_n$, to prove it let us first return to your recurrence relation:
$$\phi_{n} =\frac{-\phi_{n-1}\phi_1}{\prod_{j=1}^{n-1}(1-\phi_{j}^2)}$$
So if we tray to express $\phi_{n+1}$ using $\phi_n$ we can get the following results:
$$\begin{align}\phi_{n+1} &=&&\frac{-\phi_{n}\phi_1}{\prod_{j=1}^{n}(1-\phi_{j}^2)}\\
&=&&\frac{\phi_n}{\phi_{n-1}(1-\phi_n^2)}. \frac{-\phi_{n-1}\phi_1}{\prod_{j=1}^{n-1}(1-\phi_{j}^2)}\\
&=&&\frac{\phi_n^2}{\phi_{n-1}(1-\phi_n^2)}
\end{align}$$
So the sequence can be defined using the following relations:
$$\begin{align}
\phi_1&=&&\frac{\theta}{1+\theta^2}\\
\phi_2&=&&\frac{-\theta^2}{1+\theta^2+\theta^4}\\
\phi_{n+1}&=&&\frac{\phi_n^2}{\phi_{n-1}(1-\phi_n^2)}
\end{align}
$$
We consider $f(n)=1+\theta^2+\cdots+\theta^{2n}$. so let's prove by induction that:
$$
\phi_n=\frac{-(-\theta)^n}{f(n)} \ \ \ (*)
$$
first $(*)$ is true for $n=0,1,2...$, so assume that $(*)$ is true for $n-1$ and $n$
we have :
$$\begin{align}
\phi_{n+1}&=&&\frac{\phi_n^2}{\phi_{n-1}(1-\phi_n^2)}\\
&=&&\frac{\theta^{2n}f(n-1)}{f(n)^2.(-(-\theta)^{n-1}) (1-\frac{\theta^{2n}}{f(n)^2})}\\
&=&&-(-\theta)^{n+1}\frac{f(n-1)}{f(n)^2-\theta^{2n}}
\end{align}$$
because $-(-\theta)^{n+1}.-(-\theta)^{n-1}=\theta^{2n}$ and we have also :
$$\begin{align}f(n)^2-\theta^{2n}&=&&f(n-1)+\theta^{2n})^2-\theta^{2n}\\
&=&&f(n-1)^2+2\theta^{2n}f(n-1)+\theta^{2n}(\theta^{2n}-1)\\
&=&&f(n-1)^2+\theta^{2n}f(n-1)+\theta^{2n}(\theta^2-1)f(n-1)\\
&=&&f(n-1)(f(n-1)+2\theta^{2n}+\theta^{2(n+1)}-\theta^{2n})\\
&=&&f(n-1)f(n+1)
\end{align}$$
finally :
$$\phi_{n+1}=\frac{-(-\theta)^{n+1}}{f(n+1)} $$
