Think of ω as ω0, and let ω1 = ℵ(ω0). In other words, ω1 is the least cardinal that won't fit into ω0. Establish the recursive definition ωn+1 = ℵ(ωn), the "next" cardinal number. Here n is any ordinal, and n+1 is the successor of n.
If b is a limit ordinal, let ωb be the union of ωc for all c ∈ b. This is an ordinal, but is it a cardinal? Suppose ωb maps onto an earlier ordinal w. Remember that w is a member of a cardinal class, and ωb maps onto any set of the same size. This includes the cardinal that represents w, which we will call ωc. Let d be the successor of c, and note that the same map, from ωb onto ωc, embeds ωd into ωc. Yet ωd is, by definition, the cardinal that won't map 1-1 into ωc. Therefore ωb doesn't map onto a lower ordinal, and is a cardinal. In fact it is the least cardinal that is larger than ωc for all c ∈ b.
apply the recursion theorem and we have a function from the ordinals into the infinite cardinals. This should be called the ω() function, but it's not. Just to confuse you, this function is called ℵx, with x as subscript, which is distinctly different from the earlier function ℵ(x), where x is in parentheses. To clarify, ℵ(x) is the least cardinal that won't map into the set x; ℵx is the cardinal associated with the ordinal x, i.e. the xth infinite cardinal.
Note that ℵx is an increasing function. For x < y, ℵx < ℵy. Therefore the map is 1-1. If it is not onto, let w be the least infinite cardinal with no preimage. Either there is a largest cardinal below w or there isn't. If z is the largest cardinal below w, let ℵx = z, whence ℵx+1 = w. If w is a limit cardinal, take the unionn of the preimages of the lower cardinals to get an ordinal x, whence ℵx = w. The map is 1-1 and onto, hence ordinals and infinite cardinals correspond.
This doesn't mean they are the same size sets; they aren't sets at all. We already showed the ordinals don't form a set, hence the infinite cardinals don't form a set either.