# Proof of $LCM(a,b)=\prod_{i=1}^{\infty}p_i^{\max(\alpha_i,\beta_i)}$

For $$a,b\in\mathbb{N}$$ with prime factorization $$a=\prod_{i=1}^{\infty}p_i^{\alpha_i}, b=\prod_{i=1}^{\infty}p_i^{\beta_i}$$ with $$\alpha_i,\beta_i\in\mathbb{N}_0$$ prove: $$LCM(a,b)=\prod_{i=1}^{\infty}p_i^{\max(\alpha_i,\beta_i)}$$

I know, there is another post about this question (Prove that $lcm(a , b) = \prod_{i=1} (P_i)^{\max(\alpha_i,\beta_i)}$) but I'm wondering if there is a more detailed way to prove that the LCM of two natural numbers can be factorized in primes by their highest exponents. I don't know how to start the proof correctly.

Thanks a lot!

Obviously, don't consider the primes with exponent $$0$$ in the $$\prod_{i=1}^{\infty}$$.
$$\prod_{i=1}^{\infty} p_i^{\max(\alpha_i,\beta_i)}$$ is a common multiplier.
Suppose that $$LCM(a,b) = \prod_{i=1}^{\infty} p_i^{e_i}$$ with $$e_j < \max(\alpha_j, \beta_j) \exists j$$. WLOG $$e_j < \alpha_j$$, then $$LCM(a,b)$$ is not a multiplier of $$a$$, contradiction.