Number theory: how to prove $\gcd(a,b,c)=\prod p^{\min(a_p,b_p,c_p)}$ $a,b,c\in\mathbb Z\setminus\{0\}$
$\gcd(a,b,c):=\gcd(\gcd(a,b),c)$
How to show that $\gcd(a,b,c)=\prod p^{\min\{a_p,b_p,c_p\}}$?
My idea is to use $\gcd(a,b)=\prod p^{\min\{a_p,b_p\}}$, but I am stuck on it. Any hints?
Thanks!

The gcd of given $n$ natural numbers $a_1, a_2, \cdots ,a_n$ is defined as the largest divisor common to $a_1, a_2, \cdots ,a_n$.
If we write the prime factorization of each of the numbers $a_1, a_2, \cdots,a_n$, the gcd will be the products of some power of all the primes that are common in each number $a_1, a_2, \cdots,a_n$. The smallest power is taken into the product for gcd.
For example, let $a_1 = 72, a_2 = 60 ,a_3 = 54.$ The prime factorization is given as :
$$72 = 2^3 \cdot 3^2,$$ $$60 = 2^2 \cdot 3 \cdot 5,$$ $$54 = 2 \cdot 3^3$$
The gcd will be $$ 2^{min(3,2,1)} \cdot 3^{min(2,1,3)} \cdot5^{min(1,0)}  = 2 \cdot 3 = 6.$$
