Let $0<x\leq 1$. Prove that there exists a unique sequence $(p_n)$ of integers such that $1<p_1\leq p_2 \leq\cdots$ and $$x=\sum_{n=1}^\infty\frac{1}{p_1\dots p_n}.$$
For the existence part I completely lost, can someone give an idea?
For the uniqueness part I'm also stuck but I tried the following. Let $$x=\sum_{n=1}^\infty\frac{1}{q_1\dots q_n}$$ be another representation of $x$ such that, for $n_0\in\mathbb{N}$ such that for $n< n_0$, $q_n=p_n$ but $p_{n_0}<q_{n_0}$. Then it must be
$$ \sum_{n_0}^\infty \frac{1}{p_1\dots p_{n_0}\dots p_n} = \sum_{n_0}^\infty \frac{1}{p_1\dots p_{n_0-1}q_{n_0}q_{n_0+1}\dots q_n}. \tag{$*$} $$
With this I want to deduce the contradiction $x<x$. To do this I compute
\begin{equation} \begin{split} x &= \sum_{n=1}^{n_0-1}\frac{1}{p_1\dots p_n} + \frac{1}{p_1\dots q_{n_0}} + \sum_{n=n_0+1}^\infty\frac{1}{p_1\dots q_{n_0}\dots q_n}\\ &< \sum_{n=1}^{n_0-1}\frac{1}{p_1\dots p_n} + \frac{1}{p_1\dots p_{n_0}}+ \sum_{n=n_0+1}^\infty\frac{1}{p_1\dots q_{n_0}\dots q_n}\\ &< \sum_{n=1}^{n_0-1}\frac{1}{p_1\dots p_n} + \frac{1}{p_1\dots p_{n_0}} -\frac{1}{p_1\dots q_{n_0}} +\sum_{n=n_0}^\infty\frac{1}{p_1\dots p_{n_0}\dots p_n}\\ &= x + \frac{1}{p_1\dots p_{n_0}} -\frac{1}{p_1\dots q_{n_0}}, \end{split} \end{equation}
where in the second inequality I used $(*)$. Obviously the last part is a number greater than zero $\alpha$, so $x<x+\alpha$. This is the farthest I can get.
