Problem from Victor Prasolov's Polynomials -- Finding the number of real roots of $nx^{n}-x^{n-1}-\cdots -1$ In Chapter 1 of Polynomials by Victor Prasolov, Springer, 2001, the following theorem is proved. (p.3)

Theorem 1.1.4 (Ostrovsky). Let
$f(x)=x^{n}-b_{1}x^{n-1}-\cdots -b_{n}$,
where all the numbers $b_{i}$ are non-negative and at least one of them
is nonzero. If the greatest common
divisor of the indices of the positive
coefficients $b_{i}$ is equal to
$1$, then $f$ has a unique positive
root $p$ and the absolute value of
the other roots are $<$ p.

The following is one of the Problems to Chapter 1 (p.41).

Problem 1.5 - Find the number of real
roots of the following polynomials
a) ...
b) $nx^{n}-x^{n-1}-\cdots -1$

Question: How to solve this Problem?

Added: $nx^{n}-x^{n-1}-\cdots -1=0$ $\Leftrightarrow x^{n}-\dfrac{1}{n}x^{n-1}-\cdots -\dfrac{1}{n}=0$
Added 2: Sturm's Theorem.
 A: I think there is some value here in knowing how to do such problems "by hand."  The proof in this case is quite simple.  If $|x| > 1$, then $|x^n| > |x^k|$ for $k < n$, hence
$$n |x|^n > |x|^{n-1} + |x|^{n-2} + ... + |1| \ge |x^{n-1} + x^{n-2} + ... + 1|$$
by the triangle inequality, so this polynomial $f(x)$ has no roots of absolute value greater than $1$.  It follows that any real roots lie in $[-1, 1]$.  By inspection $x = 1$ is a root and $x = 0, -1$ are not, so any remaining roots lie in $(0, 1)$ or $(-1, 0)$.  If $x \in (0, 1)$, then
$$x^{n-1} + x^{n-2} + ... + 1 > nx^n$$
so there are no roots in $(0, 1)$.  To find any remaining roots in $(-1, 0)$, let
$$g(x) = f(x) (x - 1) = nx^n(x - 1) - x^n + 1 = nx^{n+1} - (n+1) x^n + 1.$$
Then $g'(x) = n(n+1) x^n - n(n+1) x^{n-1} = n(n+1) x^{n-1}(x - 1)$ has roots $x = 0, 1$, hence $g$ is monotonic on $(-1, 0)$, so to determine if there are roots on this interval it suffices to compute $g(-1)$ and $g(0)$.  We have $g(0) = 1$ and $g(-1) = -2n$ if $n$ is even and $g(-1) = 2n+2$ if $n$ is odd.  In the first case there is one real root in $(-1, 0)$ by the IVT and in the second case there are none.
