Here is a purely combinatorial proof of a more general identity
$$
\sum_{\lambda\vdash n}(-1)^{n-k}\frac{k!}{m_1!\cdots m_n!}\prod_{i=1}^k\binom{A}{\lambda_i}=\binom{A+n-1}n.
$$
As Ilya suggests, we rewrite this as $$\sum_{\lambda_i\geq 1, \; \lambda_1+\dots+\lambda_k=n}
(-1)^{n-k}\prod_{i=1}^k{A\choose \lambda_i}.$$
This is an alternating sum of sequences $(S_1,\dots,S_k)$ of subsets of $\{1,2,\dots,A\}$, $\sum |S_i|=n$. We identify each subset $S_i$ with an increasing sequence $p_{i1}<\dots<p_{i|S_i|}$. Then we have a sequence of length $n$: $(u_1,\dots,u_n)=(p_{11},\dots,p_{1|S_1|},p_{21}\dots,p_{2|S_2|},\dots,p_{k|S_k|})$, which we should partition onto $k$ strictly increasing pieces. If we fix a sequence $(u_1,\dots,u_n)$ and partition it onto the minimal number of strictly increasing pieces, then we may subpartition each of them in arbitrary way. This makes the corresponding alternating sum equal to 0 unless all pieces contain exactly 1 number, in other words, the sequence is non-strictly decreasing. But the number of non-strictly decreasing sequences of length $n$ and elements from 1 to $A$ equals $\binom{A+n-1}n$, this is combinations with repetition formula.