$k$ times's draw from $n$ numbers with replacement, each number at least appear once Sample $k$ times with order from $n$ distinct numbers with replacement where $k\ge n$. Here "with order" means the sample is treated like a sequence rather than a set, so results like "1,3,3" and "3,1,3" are different.
The question is what is the probability that every number is sampled at least once? My answer is $1-\frac{{\sum\limits_{i = 1}^n {C_n^i\times i! \times {{(n - i)}^k}} }}{{{n^k}}}$ where $C_n^i$ denotes $n$ chooses $i$.
${C_n^i\times i! \times {{(n - i)}^k}}$ means first choose $i$ ($i=1,2,...,n-1$) of numbers not to appear in the sample, and then we sample from the remaining $n-i$ numbers for $k$ times. 
My question is, is there more elegant solution where the numerator can be directly computed? i.e. can we directly calculate the number of possible sample sequences in which every number has at least one occurrence? Thank you!

PS: my attempt was $C_k^n\times n!\times n^{n-k}$, meaning that we first choose $n$ positions in the sequence of length $k$ to place $n$ numbers, and then permute the $n$ numbers. This is to ensure every number to appear once. The remaining $n-k$ are sampled from the $n$ numbers without restriction. 
However, I later found this seems not correct, and it counts the same sequence multiple times.
PS2: as pointed out in the comments, my first answer "1- ..." is not correct either.
 A: 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 $m=0$.
A: Hint It's straightforward to apply the Inclusion-Exclusion Principle. Alternatively, we can exploit the symmetry of the drawing process to realize this situation as a Markov chain: We define the $n + 1$ states $S_0, \ldots, S_n$ by declaring $S_i$ to be the state where precisely $i$ of the $n$ numbers have already been drawn.
By definition, if we are in state $S_i$, at the next step we have probability $\frac{i}{n}$ of remaining in that state and probability $\frac{n - i}{n}$ of advancing to the next state (in particular, state $S_n$---the state in which all numbers have been drawn at least once---is absorbing). So, the transition matrix (using the natural ordering $S_0 < \cdots < S_n$ of states) is
$$T := \pmatrix{\cdot&1&\cdot\\\cdot&\frac{1}{n}&\frac{n - 1}{n}\\\cdot&\cdot&\frac{2}{n}&\ddots\\&&\ddots&\ddots&\ddots\\&&&\ddots&\frac{n - 2}{n}&\frac{2}{n}&\cdot\\&&&&\cdot&\frac{n - 1}{n}&\frac{1}{n}\\&&&&\cdot&\cdot&1}.$$
Since the initial state of the process is $S_0$, the after $k$ draws the distribution of the states is given by the (co)vector $$E_0^\top T^k, \qquad \textrm{where} \qquad E_0 := \pmatrix{1\\0\\\vdots\\0} .$$
In particular, the probability that the system is in state $S_n$ after $k$ draws is
$$P(k) := E_0^\top T^k E_n, \qquad \textrm{where} \qquad E_n = \pmatrix{0\\\vdots\\0\\1}.$$
To find a closed form for $T^k$ and hence for $P(k)$, we use the usual trick of putting $T$ in Jordan normal form, say, $T = Q J Q^{-1}$ for appropriate matrices, $J, Q$. Then,
$$P(k) = E_0^\top (Q J Q^{-1})^k E_n = E_0^\top Q J^k Q^{-1} E_n ,$$ but $J^k$ is straightforward to compute.

The matrix is upper triangular, so its eigenvalues are its diagonal entries; they are pairwise distinct, so $T$ is diagonalizable, that is, we have $$T = Q J Q^{-1}, \qquad J := \operatorname{diag}\left(0, \frac{1}{n}, \ldots, \frac{n - 1}{n}, 1\right).$$ An inductive argument shows that we can take $Q$ to be the matrix with respective entries $$Q_{ij} = (-1)^{n - j} {{n - i}\choose j}$$ (where we index the matrix entries starting with $0$). But $J$ is diagonal, so it's straightforward to compute that $$J^k = \operatorname{diag}\left(0, \left(\frac{1}{n}\right)^k, \ldots, \left(\frac{n - 1}{n}\right)^k, 1\right).$$ Putting this all together (noting that $Q^2 = I$ makes the computation a little quicker) gives $$P(k) = \sum_{j = 0}^n (-1)^{n - j} {n \choose j} \left(\frac{j}{n}\right)^k .$$ As Henry points out in his answer, the summand is similar to the formula for the Stirling numbers of the second kind, $\left\{\begin{array}{c}k\\n\end{array}\right\}$, and we can write $P(k)$ compactly as $$P(k) = \left\{\begin{array}{c}k\\n\end{array}\right\} \frac{n!}{n^k} .$$

