How to prove that the g.c.d is equal to the prime factorization raised to the minimum of two powers for the prime factorization of $a$ and $b$ as $$a = p_1^{\alpha_1}p_2^{\alpha_2}\cdots{p_t}^{\alpha_t}$$ and $$b = p_1^{\beta_1} p_2^{\beta_2} \cdots p_s^{\beta_s}.$$ I want to prove that $d = gcd(a,b)$ is equal to $${p_1}^{\min(\alpha_1, \beta_1)}{p_2}^{\min(\alpha_2, \beta_2)}\cdots {p_r}^{\min(\alpha_r\beta_r)}$$ so I begin by proving that $d$ is a common divisor of $a$ and $b$.
$$\frac{a}{d} = \frac{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_t^{\alpha_t}}{p_1^{\min(\alpha_1\beta_1)}p_2^{\min(\alpha_2\beta_2)} \cdots {p_s}^{\min(\alpha_s\beta_s)}}$$ and
$$\frac{b}{d} = \frac{p_1^{\beta_1}p_2^{\beta_2} \cdots p_s^{\beta_s}}{p_1^{\min(\alpha_1\beta_1)}{p_2}^{\min(\alpha_2\beta_2)} \cdots p_s^{\min(\alpha_s\beta_s)}}$$
I don't know if $a$ is always divisible by $d$ in $\mathbb{Z^+}$ I'm also very new to proofs can I have some guidance?

Remember what divisibility means: $d$ divides $a$ if I can find some integer $c$ such that $dc = a$.
Let's look at one term of your $\frac{a}{d}$ fraction: $\frac{p_1^{\alpha_1}}{p_1^{min(\alpha_1, \beta_1)}}$.
Since $min(\alpha_1, \beta_1) \le \alpha_1$, this fraction has an integer value of $p_1^{\alpha_1 - min(\alpha_1, \beta_1)}$
You can then do this for every prime $p$ in your expansions of both $\frac{a}{d}$ and $\frac{b}{d}$
You're not quite done, though: you've only shown that $d$ is a common divisor of $a$ and $b$ after you're done with these steps. You still need to show that it's the greatest common divisor: this is true if a common divisor of $a$ and $b$, call it $z$, divides $d$. If every common divisor divides $d$ then you can be sure that $d$ is your greatest common divisor. I'll leave this step to you.
Let me know in a comment if you have any questions!
