Decreasing sequence Aleph's

I need to show: there does **not **exist a sequence:

such that .

Supposing there exists such a sequence:

We can observe that for any . Thus we have for all . In particular, for any we have .

In a previous excercise I had to show that (without the axiom of choice): . And since is the smallest ordinal such that we obtain: .

From this I conclude we get a strictly decreasing sequence Aleph's: Since . Thus the subsequence is a sequence such that and .

Those were some of my conclusions: However I don't see where the contradiction kicks in. Can anyone offer me some suggestions? Probably there can not exist a infinite decreasing sequence of ordinals, intuitively that might sound logical but I can't proof it.

Since the set of ordinals are wellordered, our sequence: should contain a smallest ordinal. But can't we simply define that one as ?