Take a sufficiently large $m$ ($m > 2^{3000}$) and put $x = m + \frac{1}{k-1}$. Note that
$$\left\{ \left( m + \frac{1}{k-1} \right)^n \right\} = \sum_{k_i > n} \binom{n}{i} \frac{m^{n-i}}{m^{(k-1)i}}$$
Because
$$\sum_{k_i > n} \binom{n}{i} \frac{m^{n-i}}{m^{(k-1)i}} < \frac{1}{m} \sum_{i=0}^{n} \binom{n}{i} < \frac{2^{n+1}}{m} < 1,$$
and the remaining terms of $\left(m + \frac{1}{k-1}\right)^n$ are all integers. Now if $n = kt + r$ such that $k-1 \ge r \ge 0$, we have two cases:

Case 1 $r \ne 0$.
$$\left\{ \left( m + \frac{1}{k-1} \right)^n \right\} = \sum_{k_i > n} \binom{n}{i} \frac{1}{m^{k_i-n}} \stackrel{i=t+1}{>} \frac{1}{m^{k-r}}, \\ \left\{ \left( m + \frac{1}{k-1} \right)^{n-1} \right\} = \sum_{\substack{k_i > n-1 \\ i \ge t+1}} \binom{n-1}{i} \frac{1}{m^{k_i-n+1}} < 2^n \times \frac{n}{m^{k-r+1}}, \\ \frac{1}{m^{k-r}} > 2^n \times \frac{n}{m^{k-r+1}} \implies \left\{ \left( m + \frac{1}{k-1} \right)^n \right\} > \left\{ \left( m + \frac{1}{k-1} \right)^{n-1} \right\}.$$

Case 2 $r = 0$.
$$\left\{ \left( m + \frac{1}{k-1} \right)^n \right\} = \sum_{\substack{k_i > n \\ i \ge t+1}} \binom{n}{i} \frac{1}{m^{k_i-n}} < 2^n \times \frac{n}{m^k}, \\ \left\{ \left( m + \frac{1}{k-1} \right)^{n-1} \right\} = \sum_{\substack{k_i > n-1 \\ i \ge t}} \binom{n-1}{i} \frac{1}{m^{k_i-n+1}} \ge \frac{1}{m}, \\ \frac{1}{m} \stackrel{k \ge 2}{>} 2^n \times \frac{n}{m^k} \implies \left\{ \left( m + \frac{1}{k-1} \right)^{n-1} \right\} > \left\{ \left( m + \frac{1}{k-1} \right)^n \right\}.$$
Therefore,
$$\left\{ \left( m + \frac{1}{k-1} \right)^{n-1} \right\} > \left\{ \left( m + \frac{1}{k-1} \right)^n \right\} \iff k \mid n, \\ \{x^{n-1}\} > \{x^n\} \iff k \mid n.$$