Alternative Definition of Ordinal

Theorem
Let $A$ be a class, either a set or proper class. Then $A$ is an ordinal iff $A$ is well-ordered by $\Epsilon$ (the epsilon relation) and $A$ is transitive.

Necessary Condition
Suppose $A$ is an ordinal.

By Every Ordinal is a Transitive Class, $A$ is a transitive class

By Ordinal is Well-Ordered by Epsilon, $A$ is well-ordered by $\Epsilon$.

Hence, the necessary condition is satisfied.

Sufficient Condition
Suppose $A$ is both well-ordered by $\Epsilon$ and a transitive class. Then, $\Epsilon$ satisfies transitivity with the members of $A$, so $\forall x \in A: x \subseteq A$. Therefore, any element $x$ is transitive.

Because $A$ is transitive, $\forall x \in A x \subseteq A$, and since $x \subseteq A$, $x$ is well-ordered by $\epsilon$ because of Well-Ordered Subset.

Therefore, $y \in x \implies ( y \in A \land y \subset x )$ (by transitivity of $x$). Conversely, suppose $( y \in A \land y \subset x )$. $y \in A$, so $y$ is well-ordered by $\Epsilon$ and a transitive class.

$y \subset x$, so $( y \setminus x ) \not = \varnothing$. By the axiom of foundation, $\exists z \in ( y \setminus x ): \forall w \in ( y \setminus x ): w \not \in z$. $z \not \in x$, so $z = x \lor x \in z$. In either case, $x \in y$ because of $z \in y$ and either the transitivity of $\Epsilon$ or the substitutivity of equality. All together, this means that $( y \in x \iff ( y \in A \land y \subset x ) )$, for all $y$ and $x$ in $A$.

Therefore, the segment $\{ y \in A : y \subset x \} = x$

Hence, by the definition of an ordinal, $A$ is an ordinal.

Remark
Notice that within this definition, $\Epsilon$ is a foundational relation on $A$, thus eliminating the need for the axiom of regularity for our development of ordinals.

Source

 * :$7.2$