Why isn't profinite cohomology for $\Bbb{Z}_p$ cyclic? Context and my work:
Let $\Gamma=\Bbb{Z}_p\newcommand\ZZ{\Bbb{Z}}$ be the $p$-adic integers. Let $A$ be a discrete $\Bbb{Z}_p$ module. Let $\gamma=1\in\Bbb{Z}_p$. Let $\Gamma_n = p^n\Gamma$.
Then either by definition or proof, we have
$$H^i(\Gamma,A) = \newcommand\fcolim{\varinjlim\limits}\fcolim_n H^i(\ZZ/p^n\ZZ,A^{\Gamma_n}),$$
but $\ZZ/p^n\ZZ$ is a finite, cyclic group, so on the right hand side,
we should have isomorphisms 
$$H^i(\ZZ/p^n\ZZ,A^{\Gamma_n})\cong H^{i+2}(\ZZ/p^n\ZZ,A^{\Gamma_n}),$$
for $i>0$.
The only reason these would fail to induce an isomorphism on the left hand 
side is if they weren't natural, i.e., they didn't commute with the inflation maps in the direct limit.
Thus I must conclude that these isomorphisms shouldn't be natural, since $H^1(\Gamma,A) \cong A/(\gamma-1)A$,
and I'm told that $H^n(\Gamma,A) = 0$ for $n\ge 2$ (specifically, I'm told that $\ZZ_p$ is free pro-$p$, and free pro-$p$ groups have $p$-cohomological dimension 1).
The inflation maps $H^i(\Gamma/\Gamma_m,A^{\Gamma_m})\to H^i(\Gamma/\Gamma_n,A^{\Gamma_n})$ for $m \le n$ are defined 
as the induced maps of the map of pairs $(\Gamma/\Gamma_m,A^{\Gamma_m})\to (\Gamma/\Gamma_n,A^{\Gamma_n})$.
For the usual resolution of cyclic groups, this corresponds to the diagram
$$
\require{AMScd}
\begin{CD}
\cdots  
@>N_n >> 
\ZZ[\ZZ/p^n] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^n] 
@>N_n >> 
\ZZ[\ZZ/p^n] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^n] 
@>\epsilon >> \ZZ
\\
@.
@VVV
@VVV
@VVV
@VVV
@VVV\\
\cdots 
@>N_m >> 
\ZZ[\ZZ/p^m] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^m] 
@>N_m >> 
\ZZ[\ZZ/p^m] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^m] 
@>\epsilon >> \ZZ,
\end{CD}
$$
where the maps $N_n$ are the norm maps, $N_n = \sum_{i=0}^{p^n-1} \gamma^i$.
Dropping the augmentation maps $\epsilon$, and taking $\newcommand\Hom{\operatorname{Hom}}\Hom(-,A^{\Gamma_m})$
or $\Hom(-,A^{\Gamma_n})$ as appropriate, we get
$$
\begin{CD}
\cdots 
@<N_m << 
A^{\Gamma_m}
@<\gamma -1 << 
A^{\Gamma_m} 
@<N_m << 
A^{\Gamma_m}
@<\gamma -1 << 
A^{\Gamma_m}
\\
@.
@VVV
@VVV
@VVV
@VVV\\
\cdots  
@<N_n <<
A^{\Gamma_n}
@<\gamma -1 <<
A^{\Gamma_n}
@<N_n <<
A^{\Gamma_n} 
@<\gamma -1 <<
A^{\Gamma_n} .
\end{CD}
$$
We should be careful about the vertical maps, which are all identical.
The vertical map sends $\phi_a : \ZZ[\ZZ/p^m]\to A^{\Gamma_m}$, where $\phi_a(\gamma)= a\in \Gamma^m$, to the map $\tilde{\phi_a}=\iota \circ \phi_a \circ \ZZ[q]$, where $\iota$ is the appropriate inclusion and $q$ the appropriate quotient.
Then $\tilde{\phi_a}(\gamma) = \iota(\phi_a(\ZZ[q](\gamma))) = \iota(\phi_a(\gamma))=\iota(a)$. Thus the vertical maps are all the inclusion map.
I should point out here, that for an analogous calculation of group cohomology for $\widehat{\ZZ}$ in Local Fields Serre claims that these vertical maps should be multiplication by $p^{n-m}$. 
But then every other homology group (other than the $0^{\text{th}}$) is equal, and all the vertical maps are equal, so we should have naturality of the isomorphisms with respect to the inflation maps (the vertical maps here). Something must be wrong somewhere.
Questions:

  
*
  
*Where have I gone wrong above?
  
*I haven't found a source clearly giving the cohomology groups $H^n(\Gamma,A)$. Are they indeed $H^0=A^\Gamma$, $H^1 = A/(\gamma-1)A$, $H^i = 0$ for $i>1$?
  
*What should the vertical maps be in my diagram? Did I correctly compute them, or have I made a mistake somewhere? 
  
*Is there a simpler way to derive the cohomology of $\ZZ_p$ than as a direct limit of the cohomology of cyclic groups?
  
*If this approach to computing the cohomology doesn't work, could you suggest an approach that might be more helpful?
  

Edit In an exercise after Chapter XIII, Section 1 of Local Fields Serre indicates that the vertical maps should in fact not be identical. Instead, the vertical map between $i=2h$ or $i=2h+1$th cohomology should be induced by multiplication by $(p^{m-n})^h$. I don't see why this is so, but answering this would answer all my questions.
 A: Having discovered the exercise by Serre, I have a partial answer, in the sense that I can see why the formula Serre gives for the vertical maps must be true (assuming the first two vertical maps are as I described in my question), but I'm not sure how I would get there from the definition, or what's gone wrong in my understanding of how the inflation maps are defined. A more full answer would be greatly appreciated!
The partial answer
The diagram I drew in the question doesn't commute if all the vertical maps are identical.
Let $q : \newcommand\ZZ{\Bbb{Z}} \ZZ/p^n \to \ZZ/p^m$ be the quotient map.
In the following diagram, I originally thought all the vertical maps were $\ZZ[q]$.
However, this doesn't commute, since $\ZZ[q](N_n)=p^{n-m}N_m$. Thus the diagram must 
have the maps
$$
\require{AMScd}
\begin{CD}
\cdots  
@>N_n >> 
\ZZ[\ZZ/p^n] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^n] 
@>N_n >> 
\ZZ[\ZZ/p^n] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^n] 
@>\epsilon >> \ZZ
\\
@.
@V p^{n-m}\ZZ[q] VV
@Vp^{n-m}\ZZ[q] VV
@V\ZZ[q] VV
@V\ZZ[q] VV
@|\\
\cdots 
@>N_m >> 
\ZZ[\ZZ/p^m] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^m] 
@>N_m >> 
\ZZ[\ZZ/p^m] 
@>\gamma -1 >> 
\ZZ[\ZZ/p^m] 
@>\epsilon >> \ZZ,
\end{CD}
$$
but I'm not sure why this follows from the definition of the inflation maps.
The consequence of the modified diagram:
With the modified diagram, we see that $H^0(\Gamma, A)=A^\Gamma$,
$H^1(\Gamma, A)= A/(\gamma-1)A$ since the maps are just $\ZZ[q]$ in these slots.
For $H^i(\Gamma, A)$, if $A$ is $p$-primary, then the fact that we get $p^{n-m}$ factors tells us $H^i(\Gamma,A)=0$ in the direct limit, since everything gets killed once we go far enough out in the direct limit. For $A$ not $p$-primary, but torsion, we can probably reduce to the $p$-primary case by writing $A$ as a direct sum of its $q$-primary components, for $q$ prime (the other primary parts should be $p$-divisible and thus not contribute to the cohomology). However, for the application I have in mind, I'm only interested in the $p$-primary case, so I won't flesh that argument out fully now.
