Is there a good upper bound for $(x-1)^n-x^n$ for $x\ge 1$ and $n=2,3,4,...$? For an positive integer $n\ge 2$, is there a good upper bound for $(x-1)^n-x^n$ for $x> 1$? 
Revised question: Is it less than $-\log(x-1)$ for all $x>1$?
By Binomial theorem, $(x-1)^n-x^n=\sum_{k=1}^n\binom{n}{k}x^{n-k}(-1)^k$. Or by Mean Value Theorem, it is equal to $-n\xi^{n-1}$ for some $\xi\in(x-1,x)$. I wonder if there are good estimate of this function?

The upper bound $-n\zeta^{n-1} \le -n(x-1)^{n-1}$ is certainly less that $-\log(x-1)$ for $x \ge 2. 
You can get better upper bounds by using Taylor approximations of $f(t) = t^n$ about $t = x$ and Lagrange's remainder term.  For example:
$$
(x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)\zeta^{n-2} \le 
-nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} \, .
$$
and
$$
(x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3} \zeta^{n-3} \le 
-nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3}(x-1)^3\, .
$$
