Multiplication in $O(n\cdot \log n)$
Saying that you can multiply two polynomials of degree $n$ in time $O(n\log n)$ implies that, in particular, you can multiply degree-zero polynomials, i.e., constants, in constant time.  Clearly, therefore, the claims "multiplying $n$-bit numbers takes time $O(2^{\log^*n}n\log n)$" and "multiplying degree-$n$ polynomials takes time $O(n\log n)$" are relative to different models of computation so are not directly comparable.