Equivalent definitions of ordinals? The first definition of an ordinal number I found was that an ordinal number is the $\in$-image of a well-ordered set $(A,\lt)$. From this definition it was derived that an ordinal is just the set of all lesser ordinals under the $\in$ well-ordering, and every ordinal is a transitive set. 
Looking into other sources, I found that the $\in$ function is not even mentioned, but instead an ordinal is just defined as any transitive set of transitive sets. I see how the first definition implies the second, since any ordinal is transitive. But how does the second definition, which seems less restrictive to me for some reason, imply the first? Thank you.
 A: Okay, I cheated. This is based on these notes by William Weiss.
Let $A$ be a transitive set of transitive sets. We want to prove that $A$ is totally ordered with respect to $\epsilon$. I will assume the Axiom of Regularity.

Axiom of Regularity. For every $A$, $A\neq\emptyset$, there exists $x\in A$ such that $x\cap A = \emptyset$. 

First we prove $\epsilon$ is a total order on $A$. If $A$ is empty, there is nothing to do. So assume $A$ is nonempty.
Claim 1. If $a,b\in A$, and $a\subseteq b$, then either $a=b$ or $a\in b$.
Let
$$S = \{ b\in A\mid \text{there exists }a\in A\text{ such that }a\subseteq b\text{ and }a\neq b, a\notin b\}.$$
We want to show that $S$ is empty. If $S$ is not empty, then by the Axiom of Regularity there exists $b\in S$ such that $b\cap S=\emptyset$. Let $a$ be a witness to the fact that $b\in S$ (that is, $a\subseteq b$, $a\neq b$, and $a\notin b$.
Then $b-a$ is nonempty, so by the Axiom of Regularity there exists $x\in b-a$ such that $x\cap (b-a)=\emptyset$. Since $x\in b$ and $b$ is transitive, $x\subseteq b$; since $x\cap(b-a)=\emptyset$, then we must also have $x\subseteq a$. Since $x\in b$ but $a\notin b$, then $x\neq a$. Therefore, $a-x\neq \emptyset$. Again by Regularity there exists $y\in a-x$ such that $y\cap (a-x)=\emptyset$. Since $a$ is transitive and $y\in a$, $y\subseteq a$, and as above this implies that $y\subseteq x$ as well, since $y\cap(a-x)=\emptyset$.
Since $x\in b$, and $b\in A$, then $x\in A$. Moreover, since $b\cap S=\emptyset$, then $x\notin S$, so since $y\subseteq x$, it follows that either $y=x$ or $y\in x$. But $x\in b-a$, and $y\in a-x$, so we cannot have $y=x$, hence we conclude that $y\in x$. But $y\in a-x$, which means $y\notin x$. This contradiction arises from the assumption that $S\neq \emptyset$, so $S=\emptyset$. 
Therefore, for any $a,b\in A$, if $a\subseteq b$, then either $a=b$ or $a\in b$.
Claim 2. If $a,b\in A$ and $b\not\subseteq a$, then $a\in b$.
Suppose that $b\not\subseteq a$. Using Foundation, let $x\in b-a$ such that $x\cap (b-a)=\emptyset$. Since $x\in b$ and $b$ is transitive, $x\subseteq b$, hence $x\subseteq a$ as well. By Claim 1, either $x=a$ or $x\in a$; the latter is impossible since $x\in b-a$, so $x=a$. Therefore, $a\in b$. 
Putting the two together: let $A$ be a transitive set of transitive sets, and let $a,b\in A$. If $b\subseteq a$, then either $b=a$ or $b\in a$, by Claim 1; if $b\not\subseteq a$, then $a\in b$ by Claim 2. Either way, given any $a,b\in A$, we have that either $a\in b$, $a=b$, or $b\in a$, proving that $\epsilon$ is a strict total order on $A$.
