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Will really appreciate some guidance.

Let $p$ be a prime and $n$ be a positive number.

Then $p^a$ exactly divides $n$ if $p^a|n$, but $p^{a+1} \not \! | \; n$. We then write $p^a\|n$ if $a$ is the largest component of $p$ such that $p^a|a$.

Prove that if $p^a\|n$ and $p^b\|m$ then $p^{a+b}\|mn$.

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Write $n=p^a u$ and $m = p^b v$ where neither of $u,v$ is divisible by $p$.

Then $m n = p^{a+b} u v,$ and from this your conclusion follows immediately.

EDIT: The OP wanted more details. If $p^{a+b+1}$ were to divide $mn$, then we would have $mn=p^{a+b+1}w$ where $w$ is an integer (perhaps itself divisible by $p$.) In that case, $mn=p^{a+b}uv=p^{a+b+1}w,$ so that on cancelling $p^{a+b}$ we would have $uv=pw,$ which would imply that $p$ divides one of $u,v$, whereas we assumed neither of $u,v$ were divisible by $p$ above.

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    In my first version of my answer (before what is now in the "EDIT" part) I thought the "exactly divisible by" $p^{a+b}$ was clear since neither of $u,v$ are divisible by $p$. By his comment it seems to him it was not clear.2012-11-19
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Hint $\rm\,\ p^{a+b+1}\mid (jp^a)(kp^b)\:\Rightarrow\:p\mid jk\:\Rightarrow\: p\mid j\ \ or\ \ p\mid k,\ $ by $\rm\:p\:$ prime