I'm writing out what is essentially Looijenga's proof for my own reference. Let $Z_r = \{ F_1 = F_2 = \cdots = F_r = 0 \}$. From basic dimension theory (eg Shavarevich Theorem I.6.2.5, much easier than Macaulay's result), all components of $Z_r$ have dimension $\geq n-r$. If any of them had dimension $>n-r$, then $Z_n$ would have a component of dimension $>0$, a contradiction. So all components of $Z_r$ have dimension $n-r$.

Let $A_r = k[x_0, x_1, \ldots, x_n]/\langle f_1, f_2, \ldots, f_r \rangle$. Look at the exact sequence of graded modules: $$0 \to K_r \to A_r \overset{\cdot f_{r+1}}{\longrightarrow} A_{r}[d_r] \to A_{r+1}[d_r] \to 0$$ where $K_r$ is defined to be the kernel of the map it precedes.

If we knew unmixedness, we would know that $K_r=0$. Although we don't know that, we do know that the support of $K_r$ does not contain any irreducible component of $Z_r$. That is because, if $W \subset Z_r$ were an irreducible component of $Z_r$ on which $f_{r+1}$ was $0$, then $W$ would be an $n-r$ dimensional component of $Z_{r+1}$, contradiction.

So the Hilbert polynomial of $K_r$ has degree $\leq n-r-1$. (Of course, in reality, it is zero.) We now inductively prove that the Hilbert polynomial of $A_r$ has leading term $d_1 d_2 \cdots d_r \frac{x^{n-r}}{(n-r)!}$, as desired.