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Let $a_{1}, a_{2}, \ldots, a_{n}$ be pairwise coprime positive integers where $n \geq 2 .$ Prove that

$$\operatorname{lcm}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1} a_{2} \cdots a_{n}$$

My Attempt. Induction on $n$. If $n=2$, then $\operatorname{lcm}(a_1,a_2)=a_1a_2$. Assume holds for n, show for $n+1$:

$$\operatorname{lcm}(a_1,...,a_{n+1})=\operatorname{lcm}(\operatorname{lcm}(a_1,...,a_n),a_{n+1})=\operatorname{lcm}(a_1...a_n,a_{n+1})$$

If $a_1...a_n$ and $a_{n+1}$ are relatively prime, then so we are done, ıf not how should I do? Can you help? Can you add an answer as different method?

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    $\begingroup$ Since $\,a_{n+1}\,$ is coprime to $\,a_{n},\ldots,a_1\,$ is it coprime to their product by Euclid's Lemma. That concludes your proof. For another method you can use unique prime factorization. $\endgroup$ Commented Feb 20, 2020 at 1:31

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@James Ensor, you can use the property of prime numbers.

Assume that a prime number $p_i$ is a prime divisor of $a_k$, and $v_{p_k}(a_k)=q$.

Because the numbers are pairwise co-prime, therefore, $p_i$ is a prime divisor of $a_k$ and only $a_k$, and therefore $v_{p_k}(LCM(a_1,a_2,...,a_m))=q$

This implies that $LCM(a_1,a_2,...,a_m)$ is divisible by $a_1 \times a_2\times ...\times a_n$. But by definition of LCM, it is the least non-zero number to be divisible by $a_1$ and $a_2$ and $a_3$ and ... and $a_n$. Therefore, we have Q.E.D

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