Show that $\Bbb{P}(A_{i_1}\cap \cdots \cap A_{i_k})=\sum_{j=0}^k\binom{k}{j}(-1)^j\left(1-\frac{j}n\right)^N$ for every $k=0,1,\ldots,n$ 
Let $N$ distinguishable balls be distributed in $n$ distinguishable boxes. Let $A_i = \{\text{The $i$ box is not empty}\}$.
Show that $\Bbb{P}(A_{i_1}\cap\cdots\cap A_{i_k})=\sum_{j=0}^k \binom{k}{j}(-1)^j \left(1-\frac{j}{n}\right)^N$ for every $k=0,1,\ldots,n$

Ok so Im quite messed up in trying to show this. Hints?
 A: First let’s rewrite the righthand side in a more useful form:
$$\begin{align*}
\sum_{j=0}^k\binom{k}j(-1)^j\left(1-\frac{j}n\right)^N&=\sum_{j=0}^k\binom{k}j(-1)^j\left(\frac{n-j}n\right)^N\\
&=\sum_{j=0}^k\binom{k}j(-1)^j\frac{(n-j)^N}{n^N}\\
&=\frac1{n^N}\sum_{j=0}^k\binom{k}j(-1)^j(n-j)^N\;.
\end{align*}$$
Notice that $n^N$ is the total number of ways of distributing $N$ distinguishable balls amongst $n$ distinguishable boxes, so if this sum really is $\Bbb P(A_{i_1}\cap\ldots\cap A_{i_k})$,
$$\sum_{j=0}^k\binom{k}j(-1)^j(n-j)^N\tag{1}$$
must be the number of distributions in which none of the boxes $A_{i_1},\ldots,A_{i_k}$ is empty. Verifying this is an inclusion-exclusion argument. 
For $\ell=1,\ldots,k$ let $B_i$ be the set of distributions in which box $A_{i_\ell}$ is empty; it’s not hard to see that $|B_\ell|=(n-1)^N$ for each $\ell$, since each of the $N$ balls can go in any of the $n-1$ boxes different from $A_{i_\ell}$. If $1\le\ell<\ell'\le k$, $|B_\ell\cap B_{\ell'}|=(n-2)^N$, since each of the $N$ balls can go in any of the $n-2$ boxes other than $A_{i_\ell}$ and $A_{i_{\ell'}}$. And so on: the intersection of any $j$ of the sets $B_\ell$ must have cardinality $(n-j)^N$, since each of the $N$ balls can go in any of the remaining $n-j$ boxes. Now just apply the formula at the end of the Statement section of the Wikipedia article to which I linked above, what it calls the complementary form.
More intuitively, we start with all $n^N$ distributions when $j=0$. When $j=1$ we subtract the distributions in $B_1$, the distributions in $B_2$, and so on, throwing out each distribution that misses one of our $k$ chosen boxes. However, any distribution that misses two of them gets thrown out twice and must be added back in; there are $\binom{k}2$ pairs of chosen sets, and for each of them we add back in the $(n-2)^N$ distributions that were subtracted twice, once for each of those two sets. But now it turns out that distributions that left three of the $k$ chosen boxes empty have been counted once in the $j=0$ term, subtracted $3$ times in the $j=1$ term, and added back in $\binom32=3$ times in the $j=2$ term, so they’ve actually been counted a net of one time. They aren’t supposed to be counted at all, so we use the $j=3$ term to subtract them. And so on.
A: The combinatorial species here is
$$\mathfrak{S}_{=k}(\mathfrak{P}_{\ge 1}(\mathcal{Z}))
\mathfrak{S}_{=n-k}(\mathfrak{P}(\mathcal{Z}))$$
which yields the exponential generating function
$$G(z) = (\exp(z)-1)^k \exp(z)^{n-k}$$
from which coefficient extraction produces the probability
$$n^{-N} k! \sum_{q=0}^N {N\choose q} {N-q\brace k} (n-k)^q.$$
We use EGFs since we are working in a labeled context.
Extracting coefficients a different way we obtain
$$n^{-N} \times N! [z^N] G(z)
= n^{-N} \times N! [z^N] 
\sum_{j=0}^k {k\choose j} (-1)^{k-j} \exp((j+n-k)z)
\\ = n^{-N} \sum_{j=0}^k {k\choose j} (-1)^{k-j} (n-(k-j))^N
\\ = n^{-N} \sum_{j=0}^k {k\choose k-j} (-1)^{j} (n-j)^N
\\ = \sum_{j=0}^k {k\choose j} (-1)^{j} \left(1-\frac{j}{n}\right)^N.$$
The first formula is based on the observation that
when we  multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the  product of  the two generating  functions is  the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
We have also used the fact that
$${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}$$
which results from the species
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z})).$$
