Probability formula related to distribution balls in boxes. Problem
Suppose there is a distribution of $N$ distinct balls in $n$ different boxes such that each ball has the same probability to be in any box. Let $A_i=\{\text{the i-th box is not empty}\}$. 
(1) Show for all $k=1,...,n$, we have $$V_k=P(A_{i_1} \cap ... \cap A_{ik})=\sum_{j=0}^k {k \choose j}(-1)^j(1-\frac{j}{n})^N.$$
(2) Prove that if $\lim_{n,N \to \infty} \dfrac{N}{n}=\lambda$, then $\lim_{n,N \to \infty} V_k=(1-e^{-\alpha})^k$.
I am trying to show this formula by induction. For $n=2$, we have that $$P(A_{i_1} \cap A_{i_2})=1-P((A_{i_1} \cap A_{i_2})^c)=1-(P({A_{i_1}}^c)+P({A_{i_2}}^c)-P({A_{i_1}}^c \cap {A_{i_2}}^c)).$$
This last expression equals to $$1-2\frac{(n-1)^N}{n^N}+\frac{(n-2)^N}{n^N}.$$
By a simple calculation, one can verify $$\sum_{j=0}^k {k \choose j}(-1)^j(1-\frac{j}{n})^N=1-2\frac{(n-1)^N}{n^N}+\frac{(n-2)^N}{n^N}.$$
So the hypothesis is true for $n=2$. Now, I've supposed that the statement is true for $k \leq n$ and I've tried to prove it for $k=n+1$ but I couldn't. I would appreciate if someone could help me to finish the problem with an answer or with suggestions and also any hint to do part (2).
 A: $(1)$ Rather than using induction, use the Inclusion-Exclusion Principle:
\begin{eqnarray*}
P\left(\bigcap A_{i_k} \right) &=& 1 - \sum_{r=1}^{k}{P\left(A_{i_r}^c\right)} + \sum_{r \lt s}{P\left(A_{i_r}^c \cap A_{i_s}^c\right)} - \ldots + (-1)^k P(A_{i_1}^c \cap \ldots \cap A_{i_k}^c)
\end{eqnarray*}
The general term on the RHS, with an intersection of $j$ sets, is a sum of $\binom{k}{j}$ terms of the form:
\begin{eqnarray*}
P\left(\bigcap_{r=1}^{j} A_{i_r}^c \right) &=& P(\mbox{all $N$ balls are in the other $n-j$ boxes}) \\
&=& \left(1 - \frac{j}{n}\right)^N
\end{eqnarray*}
This gives us the required answer:
\begin{eqnarray*}
P\left(\bigcap A_{i_k} \right) &=& \sum_{j=0}^k {k \choose j}(-1)^j(1-\frac{j}{n})^N
\end{eqnarray*}
$(2)$ Firstly,
$$\lim_{n,N \to \infty}{(1-\frac{j}{n})^N} = e^{-jN/n} = e^{-\lambda j}$$
using the known limit (with $k=-j, \,\,x=n, \,\,m=N/n$):
$$\lim_{x \to \infty}{(1+\frac{k}{x})^{mx}} = e^{mk}$$which is the first limit in "Notable special limits" section.
Therefore,
\begin{eqnarray*}
\lim_{n,N \to \infty}{V_k} &=& \lim_{n,N \to \infty}{\sum_{j=0}^k {k \choose j}(-1)^j(1-\frac{j}{n})^N} \\
&=& \sum_{j=0}^k {k \choose j}(-1)^j\lim_{n,N \to \infty}{(1-\frac{j}{n})^N} \\
&=& \sum_{j=0}^k {k \choose j}(-1)^j e^{-\lambda j} \\
&=& \sum_{j=0}^k {k \choose j}1^{k-j} \left(-e^{-\lambda}\right)^j \\
&=& (1 - e^{-\lambda})^k \qquad\mbox{(by the Binomial Thm)}
\end{eqnarray*}
