I want to proof that the Legendre symbol is multiplicative. Therefore I use the Criterion by Euler whih claims, that for an integer $a$ and an odd prime $p$ it holds $$ \big( \frac{a}{q} \big) \equiv a^{\frac{p-1}{2}} (\bmod p). $$
My theorem: Let $a$ be an integer with it's prime decomposition $$ a = \prod\limits_{p\in\mathbb P}p^{\alpha_p}. $$ Furthermore, let $q$ be an odd prime. Then it holds $$ \big( \frac{a}{q} \big) = \prod\limits_{p\in\mathbb P}\big( \frac{p}{q} \big) ^{\alpha_p}. $$ Proof: We have $$ \big( \frac{a}{q} \big) \equiv a^{\frac{q-1}{2}} \equiv \big( \prod\limits_{p\in\mathbb P}p^{\alpha_p} \big) ^{\frac{q-1}{2}} \equiv \prod\limits_{p\in\mathbb P}p^{\alpha_p\cdot\frac{q-1}{2}} \equiv \prod\limits_{p\in\mathbb P}p^{\frac{q-1}{2}\cdot\alpha_p} \equiv \big( \prod\limits_{p\in\mathbb P}p^{\frac{q-1}{2}} \big) ^{\alpha_p}(\bmod q), $$ which follows by the known properties of products. Again by the Euler criterion it follows that $$ \big( \prod\limits_{p\in\mathbb P}p^{\frac{q-1}{2}} \big) ^{\alpha_p}(\bmod q) \equiv \big( \prod\limits_{p\in\mathbb P}\big( \frac{p}{q}\big)\big) ^{\alpha_p}(\bmod q) $$ and finally $$ \big( \prod\limits_{p\in\mathbb P}\big( \frac{p}{q}\big)\big) ^{\alpha_p}(\bmod q) \equiv \prod\limits_{p\in\mathbb P} \big(\frac{p}{q}\big) ^{\alpha_p}(\bmod q). $$
Do I miss something? And I wonder whether I know may write $$ \prod\limits_{p\in\mathbb P} \big(\frac{p}{q}\big) ^{\alpha_p}(\bmod q) = \prod\limits_{p\in\mathbb P} \big(\frac{p}{q}\big) ^{\alpha_p}, $$ because I "leave" the congruence here for an equality.