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?

You have the right idea but perhaps a nicer way to put it would the this.
The $\gcd(a,b)$ has a prime factorization.  Let $p$ be a prime factor of $gcd(a,b)$.  Then $p$ is a prime factor of both $a$ and $b$.  So if we write the prime factor of $\gcd(a,b)$ as $\gcd(a,b) = \prod p_i^{c_i}$ then we can write the prime factor of $a$ as $a = \prod q_k^{d_k} \prod p_i^{\alpha_i}$ where $p_i|\gcd(a,b)$ (i.e. $p_i$ divides both $a$ and $b$) and $q_k$ does not (ie. $q_k$ divides $a$ but not $b$). And likewise we can write the prime factor of $b$ as $b = \prod r_k^{e_k}\prod p_i^{\beta_i}$ (where $r_k$ divides $b$ but not $a$ and $p_i$ divides both $a$ and $b$).
Okay... so $\gcd(a,b)|a$ so each $p_i^{c_i}|a$ so $c_i \le \alpha_i$ and likewise $p_i^{c_i}|b$ so $c_i \le \beta_i$ so $c_i \le \min(\alpha_i, \beta_i)$.  
But $\min(\alpha_i, \beta_i)\le \alpha_i$ so $p^{\min(\alpha_i, \beta_i)}$ divides $p^{\alpha_i}$ which divides $a$ so $p^{\min(\alpha_i, \beta_i)}$ divides $a$.  And likewise $\min(\alpha_i, \beta_i)\le \beta_i$ so $p^{\min(\alpha_i, \beta_i)}$ divides $p^{\beta_i}$ which divides $b$ so $p^{\min(\alpha_i, \beta_i)}$ divides $b$.  So $p^{\min(\alpha_i, \beta_i)}|\gcd(a,b)$.
So $\min(\alpha_i, \beta_i) \le c_i$.
So $c_i = \min(\alpha_i, \beta_i)$ and $\gcd(a,b) = \prod p_i^{\min(\alpha_i,\beta_i)}$.
