Existence of solutions to a nonlinear algebraic equation 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*}
 A: 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 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.
