How to tackle this question related to gcd. Suppose $a,b$ be two positive integers such that $a = {p_{1}}^{r_{1}}{p_{2}}^{r_{2}}...{p_{n}}^{r_{n}}$ and $b = {p_{1}}^{s_{1}}{p_{2}}^{s_{2}}...{p_{n}}^{s_{n}}$, where $p_{i}$ are distinct primes and $r_{i} , s_{i} \in \mathbb N_{0}$ for $i = 1,2,...,n$ where $\mathbb N_{0} = \mathbb N \cup \{0\}$. Then show that $d = {p_{1}}^{t_{1}}{p_{2}}^{t_{2}}...{p_{n}}^{t_{n}}$ is the $gcd (a,b)$ where $t_{i} = \min \{r_{i} , s_{i}\}$ for $i = 1,2,...,n$.
It is easy to see that $d|a$ and $d|b$.But I find difficulty to show the second part i.e. if $x$ be any common divisor of $a$ and $b$.Then $x|d$.How can I show this?Please help me.
Thank you in advance. 

Let $x$ be a common divisor.
Let $q$ be prime and $q|x$ then $q|a$ and $q|b$.  The only prime factors of $a$ and $b$ are $p_i$.  So $q = p_i$ for some $i$.
Let $k$ be the highest power of $q=p_i$ that divides $x$.  That is $q^k|x$ but $q^{k+1} \not \mid x$.  Then $q^k|a$ and $q^k|b$.  If $k > r_i$ then $p_i^k\not \mid a$.  If $k > s_i$ then $p_i^k \not \mid b$.  So $k \le t_i = \min(r_i,s_i)$.
Let $x = \prod q_l^{k_l}$ be the prime factorization.  We have shown that each $q_l = $ some $p_i$ and that each $k_l \le t_i$. So we know $\prod q_l^{k_l}|\prod p_i^{t_i}$.  So $x|d$.  
So $d$ is gcd.
