Note that we just have to show it for the case $n=2$, and then we'll be done by induction on $n$. For $n=2$, let $a,b$ be the numbers, then w.l.o.g we choose $a$ first then replace it with $\frac{\text{lcm}(a,b)}{a}=\cfrac{\frac{ab}{\gcd(a,b)}}{a}=\frac{b}{\gcd(a,b)}$, so the left numbers on the board are $\frac{b}{\gcd(a,b)}$ and $b$. Then applying the same operation on $b$ we replace $b$ with $$\cfrac{\frac{b}{\gcd(a,b)}}{\gcd \left (b,\frac{b}{\gcd(a,b)}\right)}=1$$so this replaces $b$ with $1$ and applying the operation on the $\frac{b}{\gcd(a,b)}$ we make both of them one. So, assume that for $n-1$ numbers we can make all ones, then we add a number $x$, applying the procedure on $x$ we replace $x$ with $1$ which proves the problem statement.

We notice that in this procedure LCM of all numbers on the board does not increase. Also, it is possible to reduce this LCM by gradually removing a prime (previously dividing some numbers of board) from dividing any number on the board by following procedure 1) consider the number having the largest prime power $p^k$ (in other words having max k) as a divisor for a randomly chosen prime $p$ among all primes dividing the numbers on the board and apply the given algorithm. 2) do step one for all number which are divisible by $p^k$. Clearly the LCM decrease and finally it will be 1. So, we are done.

The strategy is to pick a prime number $p$ which divides the least possible numbers on the board and then chose to replace the number wich is divisible by the highest power of $p$(if there are two or more just pick one of them) in this way the least common multiple will decrease.