Note that $\frac{n!}{a_1!a_2!\ldots a_k!}$ counts the number of functions $f:\{1,\ldots,n\}\rightarrow \{1,\ldots,k\}$ such that there are $a_i$ elements mapping to $i$ - it can be understood as the multinomial coefficient ${n \choose a_1,a_2,\ldots,a_k}$. Then, if we fix some $k$ and sum this quantity up over all possible decompositions $a_1+\ldots+a_k=n$, we are counting the number of surjective functions from $\{1,\ldots,n\}$ to $\{1,\ldots, k\}$. A proportion of $1/k$ of these maps have the property that $f(1)=1$.
For any $n$ and $k$ let $S_{n\rightarrow k}$ be the set of surjective functions $f:\{1,\ldots,n\}\rightarrow \{1,\ldots,k\}$ such that $f(1)=1$. Observe that
$$|S_{n\rightarrow k}|=\frac{n!}k\cdot \sum_{a_1+\ldots+a_k=n}\frac{1}{a_1!a_2!\ldots a_k!}$$
Let $S_{n\rightarrow\text{even}}$ be the union of $S_{n\rightarrow k}$ over all even $k$ and $S_{n\rightarrow\text{odd}}$ be the union of $S_{n\rightarrow k}$ over all odd $k$. We claim that if $n\geq 2$ then $S_{n\rightarrow\text{even}}$ and $S_{n\rightarrow\text{odd}}$ are equally large. To do this, we define an involution on $S_{n\rightarrow\text{even}}\cup S_{n\rightarrow\text{odd}}$ that changes the parity of each function.
In particular, let $f$ be a surjection from $\{1,\ldots,n\}$ to some $\{1,\ldots, k\}$ such that $f(1)=1$. We will define a new surjection $f'$ from $\{1,\ldots,n\}$ to some other $\{1,\ldots,k'\}$ where $k'$ is either $k+1$ or $k-1$.
We do this definition in two steps: If there is some $i$ other than $1$ such that $f(i)=1$, we define
$$f'(x)=\begin{cases}1 & \text{if }x=1\\
k+1 & \text{if }x\neq 1 \text{ and }f(x)=1\\
f(x) & \text{otherwise.}
\end{cases}$$
This essentially takes the preimage of $1$ under $f$ and throws everything except $1$ to a new image. This is a surjection to $\{1,\ldots,k+1\}$.
Otherwise, if the preimage of $1$ under $f$ is just $\{1\}$, we will take the things that were mapped to $k$ and map them to $1$ instead. Formally, we define
$$f'(x)=\begin{cases}1 & \text{if }x=1 \text{ or }f(x)=k\\
f(x) & \text{otherwise.}
\end{cases}$$
We can see that $f''=f$ and that $f'$ has the opposite parity from $f$. Thus the map taking $f$ to $f'$ is a bijection between $S_{n\rightarrow\text{even}}$ and $S_{n\rightarrow\text{odd}}$. Expanding this, we can write an equation:
$$\sum_{k=1}^n (-1)^k|S_{n\rightarrow k}| = 0$$
and then, by dividing out by $n!$ and negating this and substituting in our expression for $|S_{n\rightarrow k}|$, we get the desired identity.
Best Answer
In your sum, you are distinguishing between the same collection of numbers when it occurs in different orders. So you'll have separate summands for $(a_1,a_2,a_3,a_4)=(3,1,2,1)$, $(2,3,1,1)$, $(1,1,3,2)$ etc.
Given a multiset of $k$ numbers adding to $n$ consisting of $t_1$ instances of $b_1$ up to $t_j$ instances of $b_j$, that contributes $$\frac{k!}{t_1!\cdot\cdots\cdot t_j!}$$ (a multinomial coefficient) summands to the sum, and so an overall contribution of $$\frac{1}{t_1!b_1^{t_1}\cdot\cdots\cdot t_j!b_j^{t_j}}$$ to the sum. But that $1/n!$ times the number of permutations with cycle structure $b_1^{t_1}\cdot\cdots\cdots b_j^{t_j}$. So this identity states that the total number of permutations of $n$ objects is $n!$.