Depth from first nonzero Koszul cohomology

Induct on ${\mathrm{depth}}_I(M)$. At depth $0$, prime avoidance gives an associated prime $\mathfrak{p}=\mathrm{ann}(m)$ containing $I$; hence $0\mathrm{\ne }m\in 0{:}_{M}I=H^0(K^{\bullet }(\mathbf{x};M))$. If the depth is positive, choose an $M$-regular $y\in I$. Adjoining $y$, an $R$-linear combination of the $x_i$, does not change the first nonzero Koszul cohomology degree: multiplication by $y$ on $K^{\bullet }(\mathbf{x};M)$ is null-homotopic, so its mapping cone has cohomology $H^i(K)\oplus H^{i-1}(K)$.

Reorder the enlarged sequence with $y$ first. Since $y$ is regular, its two-term Koszul cochain complex has only $M\mathrm{/}yM$ in degree $1$. Thus the first nonzero degree for the enlarged complex is one plus that for the induced Koszul complex on $M\mathrm{/}yM$; explicitly, lem-depth-quotient-by-regular-element gives ${\mathrm{depth}}_I(M\mathrm{/}yM)={\mathrm{depth}}_I(M)-1$. Induction proves the first equality. The finite free Koszul complex is self-dual: $K^i(\mathbf{x};M)\cong K_{n-i}(\mathbf{x};M)$, so $H^i\cong H_{n-i}$ and the first nonzero cohomology degree is $n$ minus the last nonzero homology degree.