I read somewhere in Godement-Jacquet's book (Zeta functions of simple algebras) the following claim:
Let $D$ be a division algebra over a number field $F$ of dimension $d^2$. For each place $v$ of $F$, write $D_v:=D\otimes_FF_v$. Then $D_v$ is isomorphic to some $M_{m_v}(D_v')$, where $D_v'$ is a division algebra over $F_v$ of dimension $d_v^2$, and $d_v m_v=d$. By general theory in algebraic number theory and simple algebras (all details can be found in Pierce's book Associative Algebras (GTM) or Weil's Basic Number Theory), for all but a finite set of places $d_v=1$, and such $v$ is called "unramified" (otherwise ramified). Now let $a_v:=(-1)^{m_v(d_v-1)}$. Then $a_v=1$ for almost all $v$. Consider the well-defined (essentially finite) product $\prod_va_v$.
Claim: from the Hasse reciprocity law we can deduce that $\prod_va_v=1$. (original writing by the authors: "easy consequence of Hasse's reciprocity")
This made me puzzled. The Hasse reciprocity law is: $$0\rightarrow Br(K)\rightarrow \oplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0.$$ Here the reciprocity law tells us: given $D$, if we consider its "invariants" (local Brauer groups are isomorphic to $\mathbb{Q}/\mathbb{Z}$ by invariant maps), then the sum of these invariants becomes zero in $\mathbb{Q}/\mathbb{Z}$. We can be more explicit by writing down the invariant map via the construction of cyclic algebras of uniformizers over unramified extensions.
However, I didn't see how this implies the above formula. The formula claims $$\sum_{v\in S}d\equiv\sum _{v\in S} m_v (=\sum _{v\in S} d/d_v) \mod{2}.$$ Well, the number $d_v$ is related to the local invariants of $D$, but it simply tells us the denominator of the class of $D_v$ in $Br(F_v)$. How can the reciprocity above imply this formula?
Any hint or suggestions would be welcome. Thanks a lot in advance!
Promoting the comment to an answer.
It turns out that the argument does not require any deeper understanding of class field theory other than the fundamental exact sequence (= Hasse reciprocity, which is surely deep enough!). The rest is about the arithmetic of rational numbers (which may be why the authors called it easy).
Let us denote the class of $[D_v]$ in $Br(K_v)\simeq \Bbb{Q}/\Bbb{Z}$ by $r_v/d_v+\Bbb{Z}$. This is non-trivial only at a finite set of places $S$. We also know that $\gcd(r_v,d_v)=1$ and $$\sum_{v\in S}\frac{r_v}{d_v}\in\Bbb{Z}.\qquad(*)$$
The claim $\prod_v a_v=1$ is about how often the integer $m_v(d_v-1)$ can be odd. For $m_v(d_v-1)$ to be odd, both factors need to be odd, so $m_v$ must be odd, and $d_v$ must be even. This never happens unless $d=d_vm_v$ is even, so we can assume that $d=2^\ell d_2$ where $\ell>0$, $2\nmid d_2$.
This means that $m_v$ is odd only when $2^\ell\mid d_v$. For all $v\in S$ we have that $2^\ell r_v/d_v$ is a $2$-adic integer. Furthermore, it is a $2$-adic unit if and only if $2^\ell\mid d_v$. That is, if and only if $m_v(d_v-1)$ is odd. By $(*)$ the sum $\sum_{v\in S}2^\ell r_v/d_v$ is an even integer, so we can conclude that $m_v(d_v-1)$ is an odd integer at an even number of places. This is exactly the claim.