Proof

By hypothesis, $p > 2$ is a prime number and $r\ge 2$, $m,n_1,\ldots,n_r$ are integers.
Obviously, if $p$ is a divisor of any of the integers, then, by definition, the corresponding Legendre symbol modulo $p$ equals $0.$
Therefore, any product of Legendre symbols involving such an integers also trivially equals $0.$
We shall therefore assume that $p$ does not divide any of the integers and look at the non-trivial cases:
Case $r=2.$
- According to the Euler's criterion for quadratic residues, the Legendre symbol modulo $p$ of the product $nm$ fulfils the following congruence modulo $p$: $$\left(\frac{nm}{p}\right)\equiv (nm)^{\frac {p-1}2}\equiv (n)^{\frac {p-1}2}\cdot (m)^{\frac {p-1}2}\equiv \left(\frac{n}{p}\right)\cdot\left(\frac{m}{p}\right)\mod p.$$
- Now, observe that the difference $\left(\frac{nm}{p}\right)-\left(\frac{n}{p}\right)\left(\frac{m}{p}\right)$ can only take one of the values $0,$ $-2,$ or $2.$
- But since $p$ is odd, the above congruence holds if and only if $$\left(\frac{nm}{p}\right)-\left(\frac{n}{p}\right)\left(\frac{m}{p}\right)=0.$$
Case $r > 2.$
- This case follows by induction from the case $r=2.$
Altogether, we have shown that the Legendre symbol modulo $p$ is a completely multiplicative arithmetic function.
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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927