With $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, both $a$ and $n$ as positive, odd integers, and defining the Jacobi symbol as (where on the right hand side we have the Legendre symbol):
$$\left( \frac{a}{n} \right) = \prod_{i=1}^k \left( \frac{a}{p_i} \right)^{e_i}$$
From this definition and the law of quadratic reciprocity for the Legendre symbol:
$$\left( \frac{a}{n} \right) = \prod_{i=1}^k \left( \frac{a}{p_i} \right)^{e_i} = \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i-1}{2}} \left( \frac{p_i}{a} \right) \right)^{e_i}$$
$$\left( \frac{a}{n} \right) = \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2}} \left( \frac{p_i}{a} \right) \right)^{e_i} = \left( \frac{n}{a} \right) \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2}} \right)^{e_i} $$
So based on the law of quadratic reciprocity, we must now have:
$$(-1)^{\frac{a - 1}{2} \cdot \frac{n - 1}{2}} = \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2}} \right)^{e_i} = \prod_{i=1}^k (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2} \cdot e_i}$$
However, I don't see how this could be true, given that the right hand side is equivalent to saying that $\frac{n-1}{2} = \sum_{i=1}^k \frac{p_i - 1}{2} e_i$, when $n$ is also $p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$. So I assume I did something wrong, but I'm not sure what that is.