Setting
There is the following statement about infinite products:
If $\prod_{n=0}^\infty a_n$ is convergent (which by convention means that the limit is not zero), then $\lim_{n\to\infty} a_n = 1$.
The common proof is to apply $\ln$ to transform the partial products into partial sums, and then to use the corresponding statement about convergent infinite series (stating that summands form a zero sequence).
Question
I'm looking for an elementary $\epsilon$-proof, without the usage of $\ln$. Also, I would like to avoid inversion of the elements $a_i$.
What I have tried
So far, I have: If $\prod_{n=0}^\infty a_n$ is convergent, then the sequence of partial products $$p_n = a_0\cdot a_1 \cdots a_n$$ is a Cauchy sequence. Hence for $\epsilon > 0$, there is an $N$ such that for all $n \geq N$, $$ \epsilon > \lvert p_{n+1} - p_n\rvert = \lvert a_1\rvert\cdot \ldots \cdot \lvert a_n\rvert \cdot \lvert a_{n+1} - 1\rvert. $$ So $$\lvert a_{n+1} - 1\rvert < \frac{\epsilon}{\lvert a_1\rvert\cdot \ldots \cdot \lvert a_n\rvert}.$$
Now the problem is that I need to bound the product $\lvert a_1\rvert\cdot \ldots \cdot \lvert a_n\rvert$ from below, but I don't see a clear way how to do it.
Addition
Anne Bauval has brought up this answer. Initially, there has been a flaw in that answer, but now that it is fixed I agree that it answers my question.