Such numbers exist. Here is a way to construct one of size close to $n$, for $n\ge 2$:

Let $a_1 = n + \frac{1}{2}$. Inductively, for each $k > 1$ let $a_k = (\lfloor a_{k-1}^k\rfloor+\frac{1}{2})^{1/k}$. Then $a = \lim_{k\rightarrow\infty} a_k$ exists and is between $n$ and $n+1$.

More precisely, we have $|a_k-a_{k-1}| = |a_k^k-a_{k-1}^k|/(\sum_{i=0}^{k-1}a_k^ia_{k-1}^{k-1-i}) < \frac{1}{2kn^{k-1}}$, so $|a-a_k| < \sum_{j>k} \frac{1}{2jn^{j-1}} < \frac{1}{kn^k}$. By the same argument, if we let $\alpha_k = \inf_{j\ge k} a_k$ then we have $|a-a_k| < \frac{1}{k\alpha_k^k}$, and since $\alpha_k > a-\frac{1}{kn^k}$ we have

$|a^k-a_k^k| < |a-a_k|k(a+\frac{1}{kn^k})^{k-1} < \frac{(a+\frac{1}{kn^k})^{k-1}}{(a-\frac{1}{kn^k})^{k}} < \frac{e^{1/n^k}}{n}.$

Thus the number $a$ we've constructed will satisfy your condition with $\epsilon = \frac{1}{2}-\frac{e^{1/n}}{n}$.

Example: For $n = 10$, we have $a = 10.51174467290...$, and the smallest fractional part of $a^k$ for $k$ from $1$ to $100$ is $0.452...$. The largest is $0.543...$.