This is a classical property of $\gcd$ but I din't find a proof as mine, so I don't know if this is duplicate. (If there's a simpler proof let me know, please).
The statement is:
Let $a,b >1$ integers and suppose that their respective prime factorizations in powers of primes are
$$a=p_1 ^{\alpha_1} \cdots p_r ^{\alpha_r},$$ $$b=p_1 ^{\beta_1} \cdots p_r ^{\beta_r},$$
where $\alpha_i, \beta_i \geq 0.$
Prove that $$\gcd(a,b)=p_1 ^{\min \{ \alpha_1, \beta_1 \} } \cdots p_r ^{\min \{ \alpha_r, \beta_r \} }.$$
My attempt:
Let $d=p_1 ^{\min \{ \alpha_1, \beta_1 \} } \cdots p_r ^{\min \{ \alpha_r, \beta_r \} }.$ Observe that for all $i \in \{1,...,r \} $, we have $p_i^{\min \{ \alpha_i, \beta_i \} } \mid p_i^{\alpha_i}$ (easy to prove). Then, by divisibility properties: $$p_1 ^{\min \{ \alpha_1, \beta_1 \} } \cdots p_r ^{\min \{ \alpha_r, \beta_r \} } \mid p_1 ^{\alpha_1} \cdots p_r ^{\alpha_r}.$$ i.e., $d \mid a$.
Similarly, $d \mid b$.
Now, consider the numbers $a'=p_1^{\alpha_1 - \min \{ \alpha_1, \beta_1 \} } \cdots p_r^{\alpha_r - \min \{ \alpha_r, \beta_r \} } $ and $b'=p_1^{\beta_1 - \min \{ \alpha_1, \beta_1 \} } \cdots p_r^{\beta_r - \min \{ \alpha_r, \beta_r \} }.$
Let $d'$ a prime common divisor of $a'$ and $b'$.
In one hand, we have that $d' \mid a' \iff d' \mid p_1^{\alpha_1 - \min \{ \alpha_1, \beta_1 \} } \cdots p_r^{\alpha_r - \min \{ \alpha_r, \beta_r \} }.$ Since $d'$ is a prime number, $d' \mid p_j^{\alpha_j - \min \{ \alpha_j, \beta_j \} }$ for some $j$. Then, $\alpha_j - \min \{ \alpha_j, \beta_j \}>0, $, hence, $\min \{ \alpha_j, \beta_j \}= \beta_j. $ Observe that $d' \mid p_j, $ and then, $d'=p_j$.
In the other hand, $p_j^{\beta_j - \min \{ \alpha_j, \beta_j \}}=1.$ Therefore, $p_j \nmid b'$, since $p_j$ is not a prime factor of $b'$.
We clonclude that $a', b'$ have not common prime divisors, and then, $\gcd (a', b')=1$.
There exists $x,y \in \mathbb{Z}$ such that $$a'x+b'y=1 \hspace{1cm} (1)$$
Note that $a=da', b=db',$ then, by the equation $(1), ax+by=d.$
Then, $d \mid a, d \mid b,$ and $d$ is a linear combination of $a$ and $b$, so $d$ must be $\gcd (a,b)$