There is no infinite descending chain of ordinals

Suppose we have a decreasing sequence of ordinals $\alpha_0>\alpha_1>. . $. Recall that this means that $\alpha_{i+1}\in\alpha_i$. Then $\alpha_0$ is not well-ordered under $\in$, since $\alpha_1, \alpha_2, . . .$ is a $\in$-descending sequence of elements of $\alpha_0$. But by definition, an ordinal is a transitive set which is well-ordered by $\in$, so this is a contradiction.

(Note that it is significantly harder to show that the ordinals are linearly ordered - that is, that given distinct ordinals $\alpha$ and $\beta$, either $\alpha\in\beta$ or $\beta\in\alpha$. But if all you want is that there are no descending sequences, then this is straightforward.)