Let the written numbers be $x_1<x_2<...<x_n$ and let $\bigg\lfloor \frac{x_{i+1}}{x_i}\bigg\rfloor =q_i$, then $\frac{x_{i+1}}{x_i}\ge q_i$. Lets assume that all $q_i$'s are distinct, then $$\prod_{i=1}^{n-1} \frac{x_{i+1}}{x_i}\ge q_1.q_2.....q_{n-1}\ge 1.2.3...(n-1)$$$$\Rightarrow \frac{x_{n}}{x_1}\ge (n-1)!$$which is a contradiction since both are smaller than $(n-1)!$.

Ok this is practically same as above but I typed it up for record-keeping purposes: WLOG let our elements be $a_1, a_2, \dots a_n$ with $a_1 < a_2 < \dots < a_n$. It follows that the set $\{ \lfloor \frac{a_2}{a_1} \rfloor, \lfloor \frac{a_3}{a_2} \rfloor, \dots , \lfloor \frac{a_n}{a_{n-1}} \rfloor \}$ must consist of distinct integers. Then $ (n-1)! \leq \prod_{i=2}^{n} \lfloor \frac{a_i}{a_{i-1}} \rfloor \leq \prod_{i=2}^{n} \frac{a_{i}}{a_{i-1}} \leq a_n$, a contradiction.