Given an ideal $I\subset R=K[x_1, ...,x_n]$ and let $<$ be a term order on the ring $R$. I must show that $\forall x^u\in\operatorname{in}_<(I)$, $\exists f\in I$ s.t. $\operatorname{in}_<(f)=x^u$, where $u\in\mathbb{N}^n$ and $x^u=x^{(u_1,...,u_n)}:=x_1^{u_1}\cdots x_n^{u_n}$.
At first sight it looks like this follows directly from the definition of $\operatorname{in}_<(I):=\{\operatorname{in}_<(f):f\in I\}$, since $\forall x^u\in in_<(I),\space x^u = \operatorname{in}_<(f)$ for some $f\in I$.
However, I'm then given the hint to use the fact that $\operatorname{in}_<(fg)=\operatorname{in}_<(f)\operatorname{in}_<(g)$.
I'm unsure about what this question is asking, I've read the question countless times and it still seems to me like it's directly from the definition, but then how would the hint be useful?
As you already told, $\operatorname{in}_{<}(I)$ is the ideal generated by $\operatorname{in}_{<}(f)$ for all $f\in I$. This means that an element of $\operatorname{in}_{<}(I)$ is a linear combination of the form $g_1\operatorname{in}_{<}(f_1)+\cdots+g_r\operatorname{in}_{<}(f_r)$ with $f_i\in I$ and $g_i\in K[x_1,\dots,x_n]$. If a monomial $x^u\in\operatorname{in}_{<}(I)$ it writes $x^u=g_1\operatorname{in}_{<}(f_1)+\cdots+g_r\operatorname{in}_{<}(f_r)$. Now look at the initial monomial of the right hand side: it is of the form $\operatorname{in}_{<}(g_i)\operatorname{in}_{<}(f_i)$, for some $i$, and since $\operatorname{in}_{<}(g_if_i)=\operatorname{in}_{<}(g_i)\operatorname{in}_{<}(f_i)$ and $g_if_i\in I$ we are done.