Why is the group cohomology for a profinite group always torsion? Let $G$ be a profinite group, $A$ be a discrete $G$-module, and $n>0$ be an integer.

Why is the cohomology group $H^n(G;A)$ a torsion abelian group?

Here $H^n$ denotes the continuous cohomology groups. This thread is related, but I didn't find the answer to my question.
— I know that any (continuous) cocycle $f : G^n \to A$ has finite image, for $G$ is compact and $A$ is discrete. If the subgroup generated by the image of $f$ inside $A$ is also finite (say of cardinality $k$), then $k \cdot f = 0 : G^n \to A$ so that the class of $f$ in $H^n(G;A)$ has order at most $k$. 
If $f$ is also a group morphism, which holds if $n=1$ and $A$ is a trivial $G$-module, then the image of $f$ inside $A$ is already a subgroup, so the aforementioned condition is satisfied.
– But in general, we only want to find a multiple of $f$ which is a coboundary (without this multiple to be the zero map itself, as it was the case above). I'm not sure how to proceed. Is there may a smarter way to do it?
Thank you!
 A: $\newcommand{\Res}{\mathrm{Res}}
\newcommand{\Cor}{\mathrm{Cor}}$


*

*First of all, we can assume that $G$ is finite of order $m$. Indeed, by the proof of proposition 2.2.16 in Sharifi's notes, we deduce that for any cocycle $f : G^r \to A$ (with $r \geq 0$), the cohomology class $[f] \in H^r(G,A)$ belongs to the image of the inflation map $H^r(G/N, A^N) \to H^r(G,A)$ for some open normal subgroup $N \leq G$.
In particular, $[f]$ is a torsion element if $H^r(G/N, A^N)$ is a torsion group. Since the group $G/N$ is finite ($G$ is compact and $N$ is open), it is sufficient to prove that $H^r(G', A)$ is a torsion group whenever $G'$ is a finite group and $r>0$.

*Actually, we are going to show that $H^r(G, A)$ has finite exponent, namely its exponent is $m = |G|$.
When $r \geq 0$, the composition 
$$\Cor \circ \Res : H^r(G,A) \to H^r(\{1\}, A) \to H^r(G,A)$$
is the multiplication by $m = [G : \{1\}]$ in the cohomology group $H^r(G,M)$. This is proved in prop. II.1.30 in Milne's Class Field Theory notes ; here we are using the fact that $G$ is finite.

*Because $r > 0$, we have $H^r(\{1\}, A) = 0$. In particular, the composition $\Cor \circ \Res$ has to be the zero map.
Since it is also the multipication by $m$, this implies that $$m H^r(G,A) = 0,$$ whenever $r>0$ and $G$ is a finite group of order $m$.
This concludes the proof.
