A limit involving polynomials Let be the polynomial:
$$P_n (x)=x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1)$$
I want to prove that it has a single positive real root we'll denote by $x_n$, and then to compute:
$$\lim_{n\to\infty} x_{n}$$

Since it's not much more work, let's study the roots in $\mathbb{C}$.
Note that $x=1$ is not a solution unless $n=1$, since $P_n(1) = 1-n$.
Since we are interested in the limit $n\to\infty$, we can assume $x\ne 1$.
Sum the geometric series,
$$\begin{eqnarray*}
P_n (x) &=& x^{n+1} - (x^{n-1}+x^{n-2}+\cdots+x+1) \\
&=& x^{n+1} - \frac{x^n-1}{x-1}.
\end{eqnarray*}$$
The roots will satisfy
$$x_n^{n}(x_n^2-x_n-1) = -1.$$
(Addendum: If there are concerns about convergence of the sum, think of summing the series as a shorthand that reminds us that $(x-1)P_n(x) = x^{n}(x^2-x-1) + 1$ for all $x$.)
If $0\le |x_n|<1$, $\lim_{n\to\infty} x^n = 0$, thus, in the limit, there are no complex roots in the interior of the unit circle.
If $|x_n|>1$, $\lim_{n\to\infty} 1/x^n = 0$, thus, in the limit, the roots must satisfy 
$$x_n^2 - x_n - 1 = 0.$$
There is one solution to this quadratic equation with $|x_n|>1$, it is real and positive, 
$$x_n = \frac{1}{2}(1+\sqrt{5}).$$
This is the golden ratio. 
It is the only root exterior to the unit circle.
The rest of the roots must lie on the boundary of the unit circle. 

Figure 1. Contour plot of $|P_{15}(x+i y)|$.
