How can we prove that equation (1) has solutions for every $p$. I mean, is there an analytic method that can be used to show that there exist solutions for every $p$ for this nonlinear equation:


\begin{equation}
\left\{
\begin{array}{ccc}
x\_1^2 x\_2 \cdots x\_{p-1} x\_p+x\_1 &=& 1 \, ,\\
&&\\
 x\_1 x\_2^2 x\_3 \cdots x\_{p-1} x\_p+x\_1 x\_2 &=& 1 \, ,\\
&&\\
x\_1 x\_2 x\_3^2 x\_4 \cdots x\_{p-1} x\_p+x\_1 x\_2 x\_3&=& 1\, , \\
&&\\
\vdots & \vdots & \vdots \\
&&\\
x\_1 x\_2 x\_3 x\_4 \cdots x\_{p-1} x\_p^2+x\_1 x\_2 \cdots x\_p&=& 1\, .
\end{array}
\right.
\end{equation}


For example, let ($p=3$). Then we have:


\begin{equation}
\left\{
\begin{array}{ccc}
x\_1^2 x\_2 x\_3+x\_1 &=& 1 \, ,\\
&&\\
 x\_1 x\_2^2 x\_3 +x\_1 x\_2 &=& 1 \, ,\\
&&\\
x\_1 x\_2 x\_3^2 +x\_1 x\_2 x\_3&=& 1\, .
\end{array}
\right.
\end{equation}


So is there an analytic method for existence of solution for equation (2). For this example I used MAPLE software and got these solutions:


\begin{equation\*}
\left\{
\begin{array}{ccc}
a &=& 0.658418845314095780 \, ,\\
&&\\
b &=& 0.849466898144101812 \, ,\\
&&\\
c&=& 0.927561975482924960 \, .
\end{array}
\right.
\end{equation\*}



Existence of solutions to a nonlinear algebraic equation
Note that no $x\_k$ can vanish. If you put $ u :=x\_1x\_2\dots x\_p $ the system gives inductively, for $k=1,\dots,p$:


$$\frac{1}{x\_1 x\_2\dots x\_k}=1+u+\dots+u^k . $$


In particular $u$ solves
$$\frac{1}{u}=1+u+\dots+u^p . $$
Incidentally, $u\neq 1$, so we can express the solutions to the systems simply as
$$x\_k=\frac{u^{k}-1}{u^{k+1}-1},$$
in terms of the solutions $u\neq 1$ to the equation


$$u^{p+2}-2u+1=0. $$


Also note that, by elementary arguments, if $p$ is even the latter equation has a unique real solution (besides $u=1$), which is positive, while if $p$ is odd it has a positive and a negative solution (besides $u=1$). So the system has one or two real solutions according whether $p$ is even or odd, and in any case a unique solution 
if $x\_k$ are assumed to be real and positive.


To complete the analytic solution, we may solve $u^{p+2}-2u+1=0$ by
series using the [Lagrange inversion formula](https://en.wikipedia.org/wiki/Formal_power_series#The_Lagrange_inversion_formula) to invert $u-u^{p+2}/2$ at $1/2$: for the solution $0<u\_p<1$ one gets


$$u\_p=\frac{1}{2}\sum\_{k=0}^\infty {pk+2k\choose k} \frac{2^{-pk-2k}}{pk + k+1}.$$


For instance, 
$$u\_3=\frac{1}{2}\sum\_{k=0}^\infty {5k\choose k} \frac{32^{-k}}{4k+1}=0.5187900635...$$
that corresponds to the solutions $x\_1=a, x\_2=b, x\_3=c$ you got by MAPLE.


