Since the question is tagged "combinatorics," here's a combinatorial proof. :)
Let $A_k$ be the set of all functions $f: \{1, 2, \ldots, n\} \mapsto \{1, 2, \ldots, n+r\}$ such that $f(i) = k$ for at least one $i$, where $1 \leq k \leq n$. The identity counts the number of functions in $\cap_{k=1}^n A_k$ in two different ways.
First way: The set $\cap_{k=1}^n A_k$ is the set of onto functions from $\{1, 2, \ldots, n\}$ to itself. Since an onto function from a finite set to itself must be one-to-one as well, the intersection of the $A_k$'s is the set of bijections on $\{1, 2, \ldots, n\}$. There are $n$ choices for $f(1)$, followed by $n-1$ choices for $f(2)$ given $f(1)$, and so forth. Thus: $\text{There are $n!$ functions in } \bigcap_{k=1}^n A_k.$
Second way: Now we want to use the principle of inclusion-exclusion. The functions in the intersection of $\bar{A}_{i_1}, \bar{A}_{i_2}, \ldots, \bar{A}_{i_k}$ are those that map from $\{1, 2, \ldots, n\}$ to $\{1, 2, \ldots, n+r\} - \{i_1, i_2, \ldots i_k\}$. There are $n-k+r$ elements in this latter set, and so there are $(n-k+r)^n$ functions in $\cap_{j=1}^k \bar{A}_{i_j}$. In addition, there are $n^n$ total functions from $\{1, 2, \ldots, n\}$ to itself. By inclusion, exclusion, then, we also have the following: $\text{There are } \sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+r)^n \text{ functions in } \bigcap_{k=1}^n A_k.$