Proof by induction with a prime number I'm having trouble wrapping my head around proofs by induction with this problem:



Prove by careful induction on $k$ that, if $p$ is a prime number dividing the product $m_1m_2...m_k$ of (ordinary) integers $m_1,...,m_k$ where $k\geqslant1$, then $p$ divides at least one of the integers $m_i, i =1,...,k.$



Trying again based on feedback



So I start with the base case, that is:

$k = 1$

If $p$ divides  $m_1m_2...m_k$, that gives us $1/p$, meaning $p|1$ which supports the hypothesis.

So the base case holds.

Now for the Inductive step, we have $k +1$ which would equate to $2$

Then we have $m_1m_2...m_k = 1*2$ and then $1*2/p$.

Since $p∣ab \implies p∣a$ or $p∣b$, we know that $p|1$ or $p|2$.

So the inductive step holds, and so does the hypothesis.
Base Case:  m$_1$...m$_k$ is m$_1$.  The hypothesis implies that p thus divides m$_1$.  Thus, p divides at least one of the integers of the product m$_1$.

Recursive Step:  Suppose that for at least one m$_k$, p divides m$_1$...m$_n$.  By Bill Dubuque's hint, p divides m$_1$...m$_n$m$_j$ where j = (n + 1) implies that p divides m$_1$...m$_n$ or m$_j$.  If it divides m$_1$...m$_n$, then by assumption for at least one m$_k$ it divides m$_1$...m$_n$.  If it divides m$_j$, then for at least one m$_k$ it divides m$_j$.  Thus, in either case for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.  Therefore, if for at least one m$_k$, p divides m$_1$...m$_n$, then for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.
Base Case:  m$_1$...m$_k$ is m$_1$.  The hypothesis implies that p thus divides m$_1$.  Thus, p divides at least one of the integers of the product m$_1$.

Recursive Step:  Suppose that for at least one m$_k$, p divides m$_1$...m$_n$.  By Bill Dubuque's hint, p divides m$_1$...m$_n$m$_j$ where j = (n + 1) implies that p divides m$_1$...m$_n$ or m$_j$.  If it divides m$_1$...m$_n$, then by assumption for at least one m$_k$ it divides m$_1$...m$_n$.  If it divides m$_j$, then for at least one m$_k$ it divides m$_j$.  Thus, in either case for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.  Therefore, if for at least one m$_k$, p divides m$_1$...m$_n$, then for at least one m$_k$, p divides m$_1$...m$_n$m$_j$.