$\aleph_\alpha=\sum_{\xi<\alpha}\aleph_\xi^+$, $\alpha\neq 0$ ordinal

Question

If we suppose that $\xi\le\alpha$, since $\aleph_\xi\le\aleph_\alpha$ and $|\alpha|\le\aleph_\alpha$, then $$\sum_{\xi\le\alpha}\aleph_\xi\le\aleph_\alpha\aleph_\alpha=\aleph_\alpha.$$

It is simple to prove that $$\aleph_\alpha\le\sum_{\xi\le\alpha}\aleph_\xi.$$

Therefore, $$\aleph_\alpha=\sum_{\xi\le\alpha}\aleph_\xi.$$

I want to prove that if $\alpha\neq 0$, then $$\aleph_\alpha=\sum_{\xi<\alpha}\aleph_\xi^+,$$

but I proved that $$\aleph_\alpha=\sum_{\xi\le\alpha}\aleph_\xi.$$

I guess the steps are similar, but I don't get it. Any hint, please?

Answer 1

(1)...$\,\forall a\in Ord\, (|a|\le \aleph_a)$ by transfinite induction.

(2)...$\,e<a\implies \aleph_e<\aleph_a\implies \aleph_e^+\le \aleph_a.$

Therefore $$\sum_{e<a}\aleph_e^+\le \sum_{e<a}\aleph_a=|a|\cdot \aleph_a\le \aleph_a \cdot \aleph_a=\aleph_a.$$ I will leave the proof of $\sum_{e<a}\aleph_e^+ \ge \aleph_a$ to you.