Proposition: Let $k$ be a field and $R=k[x_1,\dots,x_n]$ the polynomial ring with $x_i$ having degree $1$. Let $f_1,\dots,f_n$ be homogeneous elements such that $\deg(f_i)=s_i >0$ and they form an $R$-sequence. Then the length of $R/(f_1,\dots,f_n)$ is equal to $s_1 s_2 \cdots s_n$.
My approach: By considering exact sequences of the form $0 \rightarrow R/(f_1,\dots,f_m)(-s_{m+1}) \stackrel{f_{m+1}}{\rightarrow}R/(f_1,\dots,f_m) \rightarrow R/(f_1,\dots,f_{m+1}) \rightarrow 0$ we can compute the Hilbert series of $R/(f_1,\dots,f_n)$ as a function of $s_i$ and of the Hilbert series of $R$, the latter being just $1/(1-t)^n$, and we can evaluate it at $t=1$ to obtain the length. I can do the calculations for a given $n$, e.g. $n=2,3$ but i am having a hard time doing the computation for a general $n$, since the expression becomes somewhat complicated.
Question: I am wondering whether there exists an inductive proof of this result. The main obstacle of course is that we would probably have to induct on $n$, but $f_1,\dots,f_n$ contain all the intermediates $x_1,\dots,x_n$.
Remark: A direction that occurred to me is replacing one of the $f_i$ by some $x_i^{s_i}$. Then i would have to show that the new sequence is again regular and that the length of the two quotient rings is the same. But it is somewhat unclear how to proceed.
Using the following exact sequence $$0 \rightarrow R/(f_1,\dots,f_m)(-s_{m+1}) \stackrel{f_{m+1}}{\rightarrow}R/(f_1,\dots,f_m) \rightarrow R/(f_1,\dots,f_{m+1}) \rightarrow 0$$ you get $H_{R/(f_1,\dots,f_{m+1})}(t)=(1-t^{s_{m+1}})H_{R/(f_1,\dots,f_m)}(t)$.
Inductively $H_{R/(f_1,\dots,f_{m+1})}(t)=(1-t^{s_{m+1}})\cdots(1-t^{s_1})H_R(t)$.