I would like to understand a proof that I saw on the proposition "Every finite abelian group is the direct product of its Sylow subgroups".
The proof starts with: let $|G| = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ be the prime decomposition of the order of $G$, where $p_i$ are all distinct and the exponents are non-zero. By Sylow theorem exists $N_i= p_i$-Sylow subgroups for each $p_i$ prime, and since $G$ is abelian we see that the $N_i$'s are normal in $G$. Therefore, $N_1\cdots N_k\leq G$ and since $|N_1\cdots N_k| = |G|$ it follows that $G=N_1\cdots N_k$ and the proof goes on, showing after that $N_i\cap N_1\cdots N_{i-1}N_{i+1}\cdots N_k = 1$ to conclude that $G$ is the direct (internal) sum of $N_1,\cdots,N_k$.
And my question is: Why does $|N_1\cdots N_k| = |G|$ ? Does it follow from the oneness of each $p_i$ sylow subgroup, since they are normal? How? I also think that it could follow from calculations... but I only know how to calculate the order for the product of two subgroups, not for a finite arbitrary number. Is this order calculated recursiverly on the order for two subgroups?
PS: I also know that $N_i\cap N_j = 1, i\neq j$.