Your first equality is not correct. Well, it happens to be correct, but what I think you're doing there is rewriting your product as $$(\prod_{0<\alpha<\omega}\alpha)\cdot(\prod_{\omega\le\alpha<\omega_1}\alpha)$$ and then replacing the first product with $\aleph_0$. However, the first product is $2^{\aleph_0}$, not $\aleph_0$.
- The issue here is a mix-up between cardinal and ordinal multiplication and notation. The question is about the cardinality of the Cartesian product of a collection of ordinals viewed as sets, which - annoyingly - corresponds to cardinal, as opposed to ordinal, multiplication. We could more clearly, if less succinctly, ask about the cardinality of the set of functions from $\omega_1\setminus\{0\}$ to $\omega_1$ satisfying $f(\eta)<\eta$ for all $0<\eta<\omega_1$.
This winds up not making a difference in the final answer, of course - $2^{\aleph_0}\cdot (\aleph_0)^{\aleph_1}$ is the same as $\aleph_0\cdot (\aleph_0)^{\aleph_1}$ - but it is still an important issue to have clearly. In particular, it makes the final argument not quite trivial: instead of just folding one factor of $\aleph_0$ into $\aleph_1$ factors of $\aleph_0$, we have to write $$2^{\aleph_0}\cdot(\aleph_0)^{\aleph_1}=2^{\aleph_0}\cdot 2^{\aleph_1}=2^{\aleph_0+\aleph_1}=2^{\aleph_1}.$$
Besides this issue, however, your answer is correct: the "after $\omega$" component of the full product dominates the rest, and it is easier to analyze all at once.
Alternatively, we could have argued as follows: we have $$\prod_{0<\alpha<\omega_1}\alpha\le\prod_{0<\alpha<\omega_1}\omega,$$ since each $\alpha<\omega_1$ has cardinality $\le\omega$ and the cardinality of the product depends only on the cardinality of the factors. But this second product is exactly $(\aleph_0)^{\aleph_1}=2^{\aleph_1}$. And $\prod_{0<\alpha<\omega_1}\alpha\ge(\aleph_1)^{\aleph_0}$ is easy to show, so we're done.
Incidentally, it's worth noting that there's a subtle axiom of choice issue here. When we calculate this product, we're implicitly fixing - for each $\omega\le\alpha<\omega_1$ - a bijection $b_\alpha: \alpha\cong\omega$. But the existence of a family of such bijections is not provable in ZF alone. In general, "large" products don't behave well in ZF alone - indeed, the axiom of choice is really just the statement that every product of nonempty sets is nonempty (= has cardinality $>0$)!
- And yes, fine, the "cardinality" of a set is a bit trickier to define in the absence of choice - see here for some details - but meh.