The number of surjections from an $n$-element set onto a $k$-element set is $\sum_{i<k+1}(-1)^{i}\binom{k}{i}(k-i)^{n}$, read in $\mathbb{R}$ through $\iota$

Let $A$ and $B$ be finite sets, $n := \lvert A\rvert$ and $k := \lvert B\rvert$, and write

$\operatorname{Surj}(A,B) := \{\, f : f \text{ is a surjection } A \to B \,\}$

(Injection, surjection, bijection). Then $\operatorname{Surj}(A,B)$ is finite and, in $\mathbb{R}$,

$\iota\lvert \operatorname{Surj}(A,B)\rvert \;=\; \sum_{i<k+1}(-1)^{i}\,\iota\binom{k}{i}\,\iota\big((k-i)^{\,n}\big),$

where $\iota$ is the canonical natural (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), $(k-i)^{n}$ is the $\mathbb{N}$-valued power of Exponentiation of natural numbers, $m^{n}$, and its agreement with the integer power in $\mathbb{R}$, and $k-i$ is the truncated difference of Finite sums and finite products of natural numbers, $\sum_{k<n} a_k$ and $\prod_{k<n} a_k$ in $\mathbb{N}$, which is the ordinary one throughout the range $i \le k$ of the sum.

All three degenerate readings are part of the statement, and each is computed rather than stipulated.

• $n = 0$ and $k = 0$. There is exactly one function $\varnothing \to \varnothing$, the empty function, and it is a surjection, so the count is $1$. The sum has the single term $(-1)^{0}\iota\binom{0}{0}\,\iota(0^{0})$, and $0^{0} = 1$ is the base clause of Exponentiation of natural numbers, $m^{n}$, and its agreement with the integer power in $\mathbb{R}$, so the sum is $1$ too.
• $n = 0$ and $k \ge 1$. No function $\varnothing \to B$ is a surjection, so the count is $0$; and every factor $\iota\big((k-i)^{0}\big)$ is $1$, so the sum is the full alternating row sum $\sum_{i<k+1}(-1)^{i}\iota\binom{k}{i}$, which is $0$ because $k \ge 1$ ($\sum_{k<n+1}\binom{n}{k} = 2^{n}$, and $\sum_{k<n+1}(-1)^{k}\iota\!\binom{n}{k} = 0$ for $n \ge 1$, clause 2).
• $n \ge 1$ and $k = 0$. There is no function from a nonempty set to $\varnothing$, so the count is $0$; and the sum has the single term $\iota(0^{n})$, which is $0$ because $n \ge 1$ (Exponentiation of natural numbers, $m^{n}$, and its agreement with the integer power in $\mathbb{R}$, clause (a)).

• The convention $0^{0} = 1$ is load bearing exactly once, at $n = 0$ and $k = 0$, where the formula returns $\iota(0^{0})$ and the truth is that the empty function is a surjection onto the empty set. It is not a convenience: it is the base clause of the recursion defining natural exponentiation, and changing it would make the formula false at that single point.

• Why the family is indexed by $B$ and not by $A$. The sieve removes the functions that miss a value, and there is one condition per element of the codomain. This is also why the alternating sum runs to $k$ and not to $n$.

• The count is a natural number. The identity is stated in $\mathbb{R}$ because it carries signs, and $\iota$ is injective, so it pins down the natural number $\lvert\operatorname{Surj}(A,B)\rvert$ exactly.