Step 1: Recall that for $\mathbb{Z}[G]$-modules $N$: $\text{Hom}_{\mathbb{Z}[G]}(N, \text{Ind}_H^G M) \cong \text{Hom}_{\mathbb{Z}[H]}(N, M)$

This is the defining adjunction property of induction.

Step 2: Since $H^n(G, -)$ is the $n$-th derived functor of the invariants functor $\text{Inv}_G$, we have: $H^n(G, \text{Ind}_H^G M) = \text{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z}, \text{Ind}_H^G M)$

Step 3: By the adjunction property: $\text{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z}, \text{Ind}_H^G M) \cong \text{Ext}^n_{\mathbb{Z}[H]}(\mathbb{Z}, M)$

To see this, take a projective $\mathbb{Z}[G]$-resolution $P_\bullet \to \mathbb{Z}$. Then: $\text{Hom}_{\mathbb{Z}[G]}(P_\bullet, \text{Ind}_H^G M) \cong \text{Hom}_{\mathbb{Z}[H]}(P_\bullet, M)$

where on the right, $P_\bullet$ is viewed as a complex of $\mathbb{Z}[H]$-modules by restriction.

Step 4: Since $P_n$ is projective as a $\mathbb{Z}[G]$-module and $G$ is a finite-index extension of $H$ (or more generally using properties of induction), $P_n$ restricted to $\mathbb{Z}[H]$ remains projective.

Actually, this requires more care: we need to show that if $P$ is a projective $\mathbb{Z}[G]$-module, then $P$ is also projective as a $\mathbb{Z}[H]$-module. This follows from:

If $P$ is a direct summand of a free $\mathbb{Z}[G]$-module $\mathbb{Z}[G]^{\oplus I}$, then as $\mathbb{Z}[H]$-modules: $\mathbb{Z}[G] \cong \bigoplus_{g \in G/H} \mathbb{Z}[H] \cdot g$

So $\mathbb{Z}[G]$ is free (hence projective) as a $\mathbb{Z}[H]$-module. Therefore $P$ is also projective over $\mathbb{Z}[H]$.

Step 5: Taking cohomology of the cochain complex $\text{Hom}_{\mathbb{Z}[H]}(P_\bullet, M)$: $H^n(\text{Hom}_{\mathbb{Z}[H]}(P_\bullet, M)) = \text{Ext}^n_{\mathbb{Z}[H]}(\mathbb{Z}, M) = H^n(H, M)$

Conclusion: Combining all steps: $H^n(G, \text{Ind}_H^G M) = \text{Ext}^n_{\mathbb{Z}[G]}(\mathbb{Z}, \text{Ind}_H^G M) \cong \text{Ext}^n_{\mathbb{Z}[H]}(\mathbb{Z}, M) = H^n(H, M)$