Prove that if $p$ is a prime and $p ~|~ a_{1} \cdot a_2 ~\cdots~ a_n$, then $p ~|~ a_k$ for some $k$, where $1 \leq k \leq n$.

Corollary: If $p$ is a prime and $p ~|~ a_{1} \cdot a_2 ~\cdots~ a_n$, then $p ~|~ a_k$ for some $k$, where $1 \leq k \leq n$.

Proof: Suppose that $p$ is a prime and $p ~|~ a_{1} \cdot a_2 ~\cdots~ a_n$, then by Theorem, $p ~|~ a_{1}$ or $p ~|~ a_2 \cdot a_3 ~\cdots~ a_n$. If $p ~|~ a_{1}$, then we are done. If $p \nmid a_{1}$, then $p ~|~ a_2 \cdot a_3 ~\cdots~ a_n$. We use Theorem, to $p ~|~ a_2 \cdot a_3 ~\cdots~ a_n$. So we will get either $p ~|~ a_2$ or $p ~|~ a_3 \cdot a_4 ~\cdots~ a_n$. If $p ~|~ a_2$, then we are done. If $p \nmid a_2$, then $p ~|~ a_3 \cdot a_4 ~\cdots~ a_n$. We will make use of Theorem  again and again. Since $n$ is finite, use of Theorem ,  will stop after finitely many times. So we will get  $p ~|~ a_k$ for some $k$, where $1 \leq k \leq n$.


Corollary: If $p, q_{1}, q_2, ~\cdots~ q_n$ all are primes and $p ~|~ q_{1} \cdot q_2 ~\cdots~ q_n$, then $p = q_k$ for some $k$, where $1 \leq k \leq n$.

Proof:  We will get $p ~|~ q_k$ for some $k$, where $1 \leq k \leq n$ by above Corollary. Since $q_k$ is prime and $p ~|~ q_k$, $q_k$ must equal to $p$. Hence $p = q_k$ for some $k$, where $1 \leq k \leq n$.

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