How prove or disprove $\gcd(\textrm{lcm}[a_{1},a_{2},\cdots,a_{n}],a_{n+1})=\cdots$ 
Let $a_{i}, i = 1, 2, \cdots, n, n + 1$ be positive integer numbers, prove or disprove
  $$\gcd([a_{1},a_{2},\cdots,a_{n}],a_{n+1})=[\gcd(a_{1},a_{n+1}),\gcd(a_{2},a_{n+1}),\cdots,\gcd(a_{n},a_{n+1})]$$
  where $[x,y] = \textrm{lcm}[x,y]$

For example
$$a_{1}=1,a_{2}=2,a_{3}=3$$
$$\gcd(\textrm{lcm}[1,2],3)=1$$
$$\textrm{lcm}[\gcd(1,3),\gcd(2,3)]=1$$
also
$$a_{1}=6,a_{2}=9,a_{3}=15$$
then
$$\gcd([6,9],15)=\gcd(18,15)=3$$
$$[\gcd(6,15),\gcd(9,15)]=[3,3]=3$$
so
$$\gcd([6,9],15)=\gcd(18,15)=[\gcd(6,15),\gcd(9,15)]$$
so this such condition.
But for general I think is true.But I can't prove it.
 A: For a prime $p$ let $v_p(a)$ be the greatest $n\ge0$ such that $p^n\mid a$.
It's easy to see that $v_p(\gcd(a,b))=\min(v_p(a),v_p(b))$ and $v_p(\text{lcm}(a,b))=\max(v_p(a),v_p(b)).$
Therefore
\begin{align*}&v_p(([a_1,\ldots,a_n],a_{n+1}))\\
&=\min(\max(v_p(a_1),\ldots,v_p(a_n)),v_p(a_{n+1}))\\
&=\max(\min(v_p(a_1),v_p(a_{n+1})), \ldots,\min(v_p(a_n),v_p(a_{n+1})))\\
&v_p([(a_1,a_{n+1}),\ldots,(a_n,a_{n+1})])\end{align*}
So they are equal, because if $v_p(a)=v_p(b)$ for all primes $p$, then $a=b$.
A: Let $a_i=\prod_{j=1}^mp_j^{\alpha_{i,j}}$ where $p_j$ are primes, $\alpha_{i,j}$ and $m$ are nonnegative integers. The existences of these parameters are clear. From the definition of gcd and lcm, we can express
$$[a_1,\ldots,a_n]=\prod_{j=1}^m p_j^{\max_{1\le i\le n}\alpha_{i,j}}.$$
It follows that
$$
\gcd([a_1,\ldots,a_n],\,a_{n+1})=\prod_{j=1}^m p_j^{\min\{\alpha_{n+1,j},\,\max_{1\le i\le n}\alpha_{i,j}\}}.
$$
On the other hand, for $1\le j\le n$, we have
$$\gcd(a_i,a_{n+1})=\prod_{j=1}^mp_j^{\min(\alpha_{i,j},\,\alpha_{n+1,j})}.$$
It follows that
$$[\gcd(a_1,a_{n+1}),\,\ldots,\,\gcd(a_n,a_{n+1})]=\prod_{j=1}^mp_j^{\max_{1\le i\le n}\{\min(\alpha_{i,j},\,\alpha_{n+1,j})\}}.$$
Therefore, the original problem is transformed to the following one: prove that
$$
\min\{\alpha_{n+1,j},\,\max_{1\le i\le n}\alpha_{i,j}\}
=\max_{1\le i\le n}\{\min(\alpha_{i,j},\,\alpha_{n+1,j})\}
$$
for each $1\le j\le m$. We can simply it by ignoring the index $j$: write $\beta_i=\alpha_{i,j}$, to prove that
$$
\min\{\beta_{n+1},\,\max_{1\le i\le n}\beta_i\}
=\max_{1\le i\le n}\{\min(\beta_i,\,\beta_{n+1})\}.
$$
Without loss of generality, we can suppose that 
$$
\beta_1\le\cdots\le\beta_k\le \beta_{n+1}\le \beta_{k+1}\le\cdots\le\beta_n,
$$
where $0\le k\le n$.
Then the above problem becomes to prove 
$$
\min\{\beta_{n+1},\,\beta_n\}
=\max\{\beta_1,\ldots,\beta_k,\beta_{n+1}\}.
$$
Now both sides are equal to the number $\beta_{n+1}$.
This completes the proof.
