Proof. It will first be proven that, if a bijection exists between two finite sets, then the two sets have the same number of elements. Let $S$ and $T$ be finite sets and $f \colon S \to T$ be a bijection. The claim will be proven by induction on $|S|$

If $|S|=0$ then $S=\emptyset$ and $f \colon \emptyset \to T$ can only be surjective if $T=\emptyset$

Assume the statement holds for any set $S$ with $|S|=n$ Let $|S|=n+1$ Let $x_1, \dots , x_{n+1} \in S$ with $S=\{x_1, \dots , x_{n+1}\}$ Let $R=S \setminus \{x_{n+1}\}$ Then $|R|=n$

Define $g \colon R \to T \setminus \{f(x_{n+1})\}$ by $g(x)=f(x)$ Since $R \subset S$ $f(x) \in T$ for all $x \in R$ Thus, to show that $g$ is well-defined, it only needs to be verified that $f(x) \neq f(x_{n+1})$ for all $x \in R$ This follows immediately from the facts that $x_{n+1} \notin R$ and $f$ is injective. Therefore, $g$ is well-defined.

Now it need to be proven that $g$ is a bijection. The fact that $g$ is injective follows immediately from the fact that $f$ is injective. To verify that $g$ is surjective, let $y \in T \setminus \{f(x_{n+1})\}$ Since $f$ is surjective, there exists $x \in S$ with $f(x)=y$ Since $f(x)=y \neq f(x_{n+1})$ and $f$ is injective, $x \neq x_{n+1}$ Thus, $x \in R$ Hence, $g(x)=f(x)=y$ It follows that $g$ is a bijection.

By the induction hypothesis, $|R|=|T \setminus \{f(x_{n+1})\}|$ Thus, $n=|R|=|T \setminus \{f(x_{n+1})\}|=|T|-1$ Therefore, $|T|=n+1=|S|$