Proof that if $k$ is the highest factor of any positive integer $n$ such that $k<n$, then $n/k$ is prime

Question

It's straightforward to say that when $n$ is prime, $k=1$ since $k$ must be less than $n$. For the case where $n$ is not prime, I thought proving that the lowest factor of $n$ is prime would be the way to go, and that $n$ divided by its highest factor is its lowest factor and the lowest factor of any number is prime. But I am unsure how to express this formally.

Answer 1

If $\frac nk$ is not prime, then $\frac nk = ab$ where $a,b\neq 1$, meaning that $ak$ and $bk$ are both factors of $n$, smaller than $n$ and larger than $k$.

Answer 2

You are on the right track. Let $1 < p < n$ is $n$'s lowest factor (so $p = n/k$). If $p$ weren't prime, say $p = p_1p_2$ with $1 < p_1 < p$. Then as $p_1 \mid p$ and $p \mid n$ implies $p_1 \mid n$, then $1 < p_1 < p$ is a factor of $n$, contradicting the fact, that $p$ was $n$'s lowest factor.