User blog comment:Edwin Shade/The Grand List Of Transfinite Ordinals/@comment-32213734-20171130020144/@comment-30754445-20171130030806

Because the way ordinal arithmetic works:

(ε4+6)Ω5 is the limit of (ε₄+6)α for all α < Ω5

Now, for any ordinals α,β < Ω5 we have: αβ < Ω5.

So the said limit cannot be any greater than Ω5:

(ε4+6)Ω5 ≤ Ω5.

And it obviously can't be any smaller, either (since (ε4+6)α is never smaller than α).

So we've found that Ω5 is both a lower bound and an upper bound of the answer, which means that:

(ε4+6)Ω₅ = Ω5.

Indeed, it can be shown that:

(ε4+6)εᵦ = εᵦ for any εᵦ>(ε4+6) (or in other words, for any β ≥ 5).

At any rate, we now have: (ε4+6)Ω₅ = Ω5.

So the original expression you've written can be reduced to ζΩ 5, which (due to very similar reasoning) is equal to Ω5