Mapping from Set to Class of All Ordinals is Bounded Above

Theorem
Let $n$ be an ordinal.

Let $\operatorname{On}$ be the class of all ordinals.

Let $f:n \to \operatorname{On}$ be a mapping.

Then $f$ has an upper bound.

Proof
Let $I$ be the image of $f$:

By the definition of ordinal, $n$ is a set.

Thus by the Axiom of Replacement, $I$ is a set.

By the Axiom of Union, $\bigcup I$ is also a set.

Thus by Union of Subset of Ordinals is Ordinal, $\bigcup I$ is an ordinal.

But by Union Smallest, each element of $I$ is a subset of $\bigcup I$.

Thus $\bigcup I$ is an upper bound of $f$.