Numerical example for $\gcd(a,b)=\prod p_i^{\min(a_i,b_i)}$

$$\begin{align}a=600&=2^3\cdot 3^1 \cdot 5^2\\ b=54&=2^1\cdot 3^3 \cdot 5^0\\[2ex] \text{GCD}(a,b)&=2^1\cdot 3^1\cdot 5^0=6\\ \text{LCM}(a,b)&=2^3\cdot 3^3\cdot 5^2=5400\\[2ex] \text{GCD}(a,b)\cdot\text{LCM}(a,b)&=2^4\cdot 3^4\cdot 5^2=32400=ab \end{align}$$

Certainly: consider $a=48=2^4\cdot 3$ and $b=40=2^3\cdot 5$. Then we let $p_1=2, p_2=3, p_3=5$ and observe that $a_1=4, a_2=1, a_3=0$ and $b_1=3, b_2=0, b_3=1$. Your corollary then tells us that \begin{align*} \gcd(48,40) &=\gcd(a,b) \\ &= \prod p_i^{\min(a_i,b_i)} \\ &= p_1^{\min(a_1,b_1)}\cdot p_2^{\min(a_2,b_2)}\cdot p_3^{\min(a_3,b_3)} \\ &= 2^{\min(4,3)}\cdot 3^{\min(1,0)}\cdot 5^{\min(0,1)} \\ &= 2^3\cdot 3^0 \cdot 5^0 \\ &= 8, \end{align*}

Using a similar method we determine that

\begin{align*} lcm(48,40) &=lcm(a,b) \\ &= \prod p_i^{\max(a_i,b_i)} \\ &= 2^4\cdot 3^1 \cdot 5^1 \\ &= 240. \end{align*}

We can then observe that $$ab=48\cdot 40=1920=8\cdot 140=\gcd(48,40)\cdot lcm(48,40)=\gcd(a,b)\cdot lcm(a,b).$$

Let me know if anything needs clarifying (and if anyone knows the command that Latex recognises as a math operator for lcm).

For example, $540 = 2^2\cdot 3^3\cdot 5$ and $72 = 2^3\cdot 3^2$, so their GCD is $2^2\cdot 3^2 = 36$ and their LCM is $2^3\cdot 3^3\cdot 5 = 1080$. The point is that if a number $d$ is to divide both $a$ and $b$, the power to which each prime appears in the factorization of $d$ cannot exceed the mininum power to which it appears in the factorization of both $a$ and $b$. A similar analysis works for the LCM.