Is $lcm( lcm(n_1, n_2, ..., n_m), n_{m+1}) = lcm(n_1, n_2, ..., n_m, n_{m+1})$?

Question

Is $lcm( lcm(n_1, n_2, ..., n_m), n_{m+1}) = lcm(n_1, n_2, ..., n_m, n_{m+1})$?

And if so, how can we prove it ?

I thought first I open $lcm(n_1, n_2, ..., n_m)$ as $n_1*a_1 + n_2*a_2 + ... = d_n$ and then write $d_n * a'_n + n_{m+1}*a_{m+1}$, but as far as I know, I cannot conclude the result from this point of view.

Answer 1

Let lcm$(n_1,...,n_m)=a,$ lcm$(a,n_{m+1})=b,$ lcm$(n_1,...,n_{m+1})=c.$

Since $\forall i = 1,...,m,$ $n_i|a$ and $a|b,n_{m+1}|b,$ we have $\forall i,$ $n_i|b,$ so $c|b.$

Since $\forall i = 1,...,m,$ $n_i|c,$ so $a|c,$ and with $n_{m+1}|c,$ we have $b|c.$

Hence $b=c.$