59 Given that $n$ is a natural number. If
$$
x^{n+1} \geqslant \sum_{k=0}^{n-1}(n-k) C_{n+1}^{k} x^{k}
$$

Find the range of values for $x$.
$\begin{array}{lllll}59 & \end{array}$
$$
\begin{aligned}
x^{n+1}-\sum_{k=0}^{n-1}(n-k) C_{n+1}^{k} x^{k} & =x^{n+1}-\sum_{k=0}^{n-1}(n+1-k) C_{n+1}^{k} x^{k}+\sum_{k=0}^{n-1} C_{n+1}^{k} x^{k} \\
& =\sum_{k=0}^{n+1} C_{n+1}^{k} x^{k}-C_{n+1}^{n} x^{n}-\sum_{k=0}^{n-1}(n+1) C_{n}^{k} x^{k} \\
& =(1+x)^{n+1}-(n+1)(1+x)^{n} \\
& =(x-n)(x+1)^{n}
\end{aligned}
$$

From $(x-n)(x+1)^{n} \geqslant 0$, we get:
When $n$ is even, $x \geqslant n$ or $x=-1$.
When $n$ is odd, $x \geqslant n$ or $x \leqslant-1$.