Proof regarding prime numbers: THEOREM:
If a prime $p$ divides a product $a_1 \cdot \cdot \cdot a_n$, then $p$ divides at least one of its factors, $a_i$. 
This is my attempt at the proof, the book I am reading from suggests using induction on the number of factors, but I went a different direction because I couldn't get anywhere with induction:
Suppose $p\mid a_1\cdot\cdot\cdot a_n$ and $ p\nmid a_i^*$ where $a_i^* = \dfrac{a_1\cdot\cdot\cdot a_n}{a_i}$ and $a_i$ is one of the factors of the product. 
We know that, 
\begin{align*}
\gcd(p,a_i^*) = 1\\
a_i = \gcd(a_ip,a_ia_i^*)
\end{align*}
And since $p\mid a_ip,a_1\cdot\cdot\cdot a_n \Longrightarrow p\mid(a_ip,a_1\cdot\cdot\cdot a_n) \Longrightarrow p\mid a_i$. 
Has the theorem been proven? 

I got confused by your argument.  The induction follows easily from the following result.  
Claim: If $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$ (or both).
Because $p$ is prime, its only divisors are $1$ and $p$.  This means $\gcd(a,p)$ must be either $p$ or $1$.  If $\gcd(a,p) = p$, then $p$ divides $a$ and we're done.
Otherwise $\gcd(a,p) = 1$.  Then using Euclid's algorithm you get
$$ra + sp = 1$$
for integers $r$ and $s$.  Multiplying both sides by $b$ you get
$$rab + spb = b$$
The left hand side is divisible by $p$ because $p$ divides $ab$.  This shows that $b$ is divisible by $p$.
Now you can use this to prove the result for an arbritary product $a_1 \cdots a_n$, by using induction on the number of factors.  What if the theorem were true for all products with $n-1$ factors?  Hint: integer multiplication is associative.
