$k$ times's draw from $n$ numbers with replacement, each number at least appear once

probability combinatorics

I think the expression for the sum should be $\dfrac{\displaystyle\sum_{j=0}^{n} (-1)^{n-j} \binom{n}{j} j^k}{n^k}$.

The nearest you will get to a simple form of a solution to your question is then $\dfrac{S_2(k,n) \, n!}{n^k}$ where $S_2(k,n)$ represents a Stirling number of the second kind, sometimes written $\displaystyle\Bigl\lbrace{k\atop n}\Bigr\rbrace$.

You can extend this to find the probability that there are $m$ items of the $n$ unsampled, which is $\dfrac{S_2(k,n-m) \, n!}{n^k\, m!}$, and in the original question $m=0$.

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