Where I've got to so far:
Let $m(X)$ ε I be a non-zero polynomial of least degree.
For any $a(X)$ ε $\mathbb{F}$$[X]$, $a(X)m(X)$ ε I as I is a principal ideal of $\mathbb{F}$$[X]$ thus $<m(x)>$ $⊆$ I.
Let us now consider $b(X)$ ε I. By the division theorem for polynomials, $b(X) =q(x)m(x) + r(X)$ where $q(X), r(X)$ ε $\mathbb{F}$$[X]$ and $r(X) = 0$ or $deg(r(X)) < deg(m(X))$.
Observe that $r(X) = b(X) - q(X)m(X)$ ε I.
If $r(X) = 0$ then we are done: We have shown that $b(X) = q(X)m(X)$ which means $I ⊆ <m(X)>$ hence I $=$ $<m(X)>$.
If $deg(r(X)) < deg(m(X))$ then observe that $m(X)$ was chosen to have the least degree in I thus $deg(r(X)) = 0$ which means $r(X)$ is a constant polynomial.
So where I am so far is $r(X)$ = $b(X) - q(X)m(X)$ where $r(X)$ is a constant polynomial. How do deduce $b(X) = q(X)m(X)$ in this situation?