Problem
For any odd composite number $m$ with at least $2$ distinct prime factors,
prove that $\displaystyle \prod_{\substack{1\leq i\leq m \\ \text{gcd}(i,m)=1}}\left(\frac{i}{m}\right) = 1$,
where $\left(\frac{.}{.}\right)$ is the Jacobi symbol, and $\text{gcd}(i,m)=1$ means that $i$ is coprime with $m$.
Attempts
I have already proved that for any odd prime number $p$,
$\displaystyle \prod_{\substack{1\leq i\leq p^n \\ \text{gcd}(i,p^n)=1}}\left(\frac{i}{p^n}\right) =
\begin{cases}
-1 &\text{ if } p \text{ is of the form } 4k+3 \text{ and } n \text{ is odd,}\\
1 &\text{ if otherwise.}
\end{cases}
$
I was planning to factorize $m$ into prime powers, then somehow combine their Jacobi Symbols together or do induction.
However, things get complicated real quick because when $m$ changes, the numbers coprime to it changes greatly.
If $m$ is of the form $p_1^{n_1}p_2^{n_2}$, it might still be manageable, but I don't know how to handle the general case.
Also, I found that its Euler's totient function $\phi(m)$ is even, which might be related.
Any help would be greatly appreciated.