This is a usual theorem and I was going through the proof from Gruenberg: Cohomological topics in Group Theory.
Starting with a projective resolution $\dotsc \rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow {\mathbb{Z}}$, we form the $({\mathrm{Hom}}_G(P_i, A))$ and take the (co-)homology of this complex. To verify minimality take a coinduced module $A = C^{\ast} := {\mathrm{Hom}}_{\mathbb{Z}}({\mathbb{Z}}G, C)$ and to verify $H^q(G,A) = 0$ for all $q \geq 1$.
Now there is an isomorphism ${\mathrm{Hom}}_G(P_i, A) \cong {\mathrm{Hom}}_{\mathbb{Z}}(P_i, C) =: K^i$.
Here is the claim that $0 \rightarrow A^G \rightarrow K^0 \rightarrow K^1 \rightarrow \dotsc$ is exact because the original resolution splits over ${\mathbb{Z}}$.
In fact, the isomorphism yield two rows of co-chain complex with isomorphic vertical arrows. So it is only needed to verify the ${\mathrm{Hom}}_{\mathbb{Z}}(-,C)$ functor takes the projective resolution to an exact cochain complex.
(i) Now first of all, while verifying the injectivity of the first arrow and exactness at the $0$-th point I don't see much use of "projective".
(ii) Now while proving ${\mathrm{Hom}}_{\mathbb{Z}}(P_{i+1},C) \rightarrow {\mathrm{Hom}}_{\mathbb{Z}}(P_i,C) \rightarrow {\mathrm{Hom}}_{\mathbb{Z}}(P_{i-1},C)$ (by induction on i): i.e., ${\mathrm{ker}}({\overline{d_{i+1}}}) \subseteq {\mathrm{Im}}({\overline{d_i}})$
I could see for a $\varphi : P_i \rightarrow C$ with $(\varphi){\overline{d_{i+1}}} = d_{i+1} \varphi = 0$, there exists $\psi : {\mathrm{Im}}(d_i) \rightarrow C$ so that $(\psi){\overline{d_i}} = d_i \psi = \varphi$. The problem now is how to extend $\psi$ to $\psi : P_{i-1} \rightarrow C$?
Usually a homomorphism $\psi : N \rightarrow C$ does not extend to $\psi : P \rightarrow C$, while $P$ is projective and $N \leq P$.
Am I missing something trivial?