Projective Resolution of exterior algebra as a module over divided polynomial algebra

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Let $\Lambda_\mathbb{Z}[x]$ be an exterior algebra on one generator with $|x|=n$, let $\Gamma_\mathbb{Z}[x]$ be a divided polynomial algebra with $|x_k|= kn$, and suppose that $\Lambda_\mathbb{Z}[x]$ is a module over $\Gamma_\mathbb{Z}[x]$.

The multiplication on $\Gamma_\mathbb{Z}[x]$ is given by $$ x_kx_l = {k+l \choose k }x_{k+l}. $$

Question: Calculate a projective resolution for $\Lambda_\mathbb{Z}[x]$ as a module over $\Gamma_\mathbb{Z}[x]$.

Attempt

I'm attempting to answer this without any overly technical machinery (bar construction/Kozul complex etc). I will denote $\mathbb{Z}_{x_k}$ a copy of the integers generated by $x_k$.

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Any help with this would be greatly appreciated. I am only interested in the case up to $\mathbb{Z}_{x_4}$ as it is all I need for the calculation purposes.

Any help would be greatly appreciated.

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We can continue computing a minimal free resolution as you have done. If I haven't made a mistake, the answer, up through degree $4n$, is $\Gamma_\mathbb{Z}[x] \cdot \{\alpha_0, \beta_2, \gamma_3, \delta_4, \epsilon_3, \zeta_4, \eta_4, \theta_4\}$ where the degree of each basis element is $n$ times the subscript. The differentials are: \begin{align*} d(\alpha_0) &= 1 \\ d(\beta_2) &= x_2 \alpha_0 \\ d(\gamma_3) &= x_3 \alpha_0 \\ d(\delta_4) &= x_4 \alpha_0 \\ d(\epsilon_3) &= 3\gamma_3 - x_1 \beta_2 \\ d(\zeta_4) &= 6\delta_4 - x_2 \beta_2 \\ d(\eta_4) &= 4\delta_4 - x_1 \gamma_3 \\ d(\theta_4) &= x_1 \epsilon_3 + 2\zeta_4 - 3\eta_4. \end{align*}