Inductive Proof of Group with Prime Decomposition is Isomorphic to Direct Product of Cyclic Groups My lecturer set as a bonus exercise the following induction proof:
If $G$ is a finite abelian group $|G| = p_1^{n_1} \cdots p_s^{n_s}$ is the decomposition of $|G|$ into a product of distinct prime numbers, then $G \simeq G_1 \times \cdots \times G_s$ where $G_i$ is an abelian group of order $p_i^{n_i}$.
I know that I am supposed to perform induction on the order of $G$, and intuition tells me that this is a strong induction style question. The base case seems straightforward enough, after all, if $|G| = p_1^{n_1}$, then it is clearly isomorphic to itself. However, the inductive step is where I have no inkling on how to proceed! If I assume that the order of $G$ is decomposable to $k$ distinct primes and that this is isomorphic to a direct product of $k$ abelian groups of prime order, then how do I use this to prove that if I have a $k+1$-th prime, then this leads to an isomorphism with a direct product of $k + 1$ groups?
Any help is much appreciated!

We'll induct on the number of distinct prime factors in the group's order. As you've said, the base case is pretty easy. Now, suppose the statement has been proven for all abelian groups, $H$, such that $|H|$ has $n$ distinct prime factors. 
Let $G$, then, be an abelian group with $n+1$ distinct prime factors. This means that we can factor $|G|$ as $p^kj$ where $p^k$ and $j$ are coprime, and $j$ has $n$ distinct prime factors. Let $P$ be the set of all elements in $G$ whose order divides $p^k$ and let $J$ be the set of all elements in $G$ whose order divides $j$. 
It is easy to see that both $P$ and $J$ are subgroups of $G$. It is also easy to see, by Cauchy's theorem, that $|P|$ and $|J|$ must be coprime. This will be important later. 
First, each element in $G$ can be expressed as a sum of an element in $P$ and element in $J$. This is relatively easy to show. For any element, $x\in G$, we have $|G|\cdot x=0$, i.e. $(p^kj)\cdot x=0$. But this can be rewritten in a couple of ways- we can write it as $j\cdot (p^k\cdot x)=0$ which gives us that $\operatorname{ord}(p^k\cdot x)|j$.  
Similarly, $\operatorname{ord}(j\cdot x)|p^k$. So, by the definitions of $P$ and $J$, for any $x\in G$, $p^k\cdot x\in J$ and $j\cdot x\in P$.
But, since $p^k$ and $j$ are coprime we can pick $a$ and $b$ so that $ap^k+bj=1$, which means $a\cdot(p^k\cdot x)+b\cdot(j\cdot x)=(ap^k)\cdot x+(bj)\cdot x=(ap^k+bj)\cdot x=1\cdot x=x$.
By what we have already shown, $a\cdot(p^k\cdot x)\in J$ and $b\cdot(j\cdot x)\in P$, so each $x\in G$ can be written as the combination of an element in $P$ and an element in $J$.  
Also, since the order of any element in $P$ is coprime to the order of any non-zero element in $J$ (by the definitions of $P$ and $J$), we have $P\cap J=\{0\}$.  
This finally gives us $G=P\oplus J$. Given that: $|P|$ and $|J|$ are coprime, $|P|=p^s$ for some $s$ (by Cauchy's theorem), and that $|G|=|P||J|$ (because $G$ is the direct sum of $P$ and $J$), we have $|P|=p^k$ and $|J|=j$. Now, we can apply the induction hypothesis on $J$ to expand it as a direct sum of groups of prime order in distinct primes, and, with $G=P\oplus J$, we are done.   
