Ordered $k$-covers of $[n]$ An ordered $k$-cover of $[n]$, for $k\in\mathbb{P}$ and $n\in\mathbb{N}$, is a sequence $(A_1,\dots,A_k)$ of subsets of $[n]$ such that $A_1\cup\dots\cup A_k=[n]$. If $m(n,k)$ is the number of ordered $k$-covers of $[n]$, then is this formula (by the principle of inclusion and exclusion) correct?
$$m(n,k)=2^{n}-\sum_{j=1}^{k}(-1)^{k-j}\binom{k}{j}j^{n}$$
I don't know how to simplify it further.... Any help would be greatly appreciated.
 A: Given Dome's comment of 09/17/13 I interpret the question as follows: We have to count the number of $k$-tuples $(A_1,A_2,\ldots,A_k)$ of subsets $A_i\subset[n]$ having the property $\bigcup_{1\leq i\leq k}A_i=[n]$.
This means that for each number $\ell\in[n]$ we can freely decide in which of the $A_i$, $\>1\leq i\leq k$, it shall occur, under the sole condition that it occurs in at least one of the $A_i$. In other words: We have to select for each $\ell\in[n]$ a nonempty subset $J_\ell\subset[k]$. When these sets $J_\ell$ have been selected put
$$A_i:=\bigl\{\ell\in[n]\ \bigm|\ i\in J_\ell\bigr\}\qquad(1\leq i\leq k)\ .$$
There are $2^k-1$ nonempty subsets of $[k]$, and we can select  one of them independently for each $\ell\in[n]$. Therefore the total number $N$ of admissible $k$-tuples $(A_1,A_2,\ldots,A_k)$ is given by
$$N=\left(2^k-1\right)^n\ .$$
A: This is an argument using Inclusion-Exclusion, but Christian Blatter's solution is much simpler:
Let S be the set of all ordered selections of k subsets of $[n]$, and let $E_{i}$ be the set of selections which do not have the element i in the union of the subsets, for $1\le i\le n$.
Then $\displaystyle \vert \overline{E_{1}}\cap\cdots\cap\overline{E_{n}}\vert=\vert S\vert-\sum_{i}\vert E_{i}\vert+\sum_{i<j}\vert E_{i}\cap E_{j}\vert-\cdots $
$\displaystyle\;\;\;\;\;=2^{nk}-\binom{n}{1}2^{(n-1)k}+\binom{n}{2}2^{(n-2)k}-\binom{n}{3}2^{(n-3)k}+\cdots+(-1)^{n-1}\binom{n}{n-1}2^k+(-1)^n$
$\displaystyle\;\;\;\;\;=x^n-\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}-\binom{n}{3}x^{n-3}+\cdots+(-1)^{n-1}\binom{n}{n-1}x+(-1)^n$  
$\displaystyle\;\;\;\;\;=(x-1)^n=(2^k-1)^n$,  using $x=2^k$.
