I posted this a while ago, but realized that actually this works, but produces information loss. Here is the process:
Let's try to reduce a number $x$. We will be using base $\beta$ for this process.
Let $x$ be our "worst case" number.
$$ x_{10} = ((\beta-1)\sum_{n=0}^k \beta^n)_{10} $$
Next thing we will do is take the number written in base $\beta$. Therefore:
$$ x_{10} = (\beta\beta..\beta)_{\beta} $$
Now we take the number in base beta and take the values literally, transforming them to the numeration system of base 10.
$$ y_{10} = ((\beta-1)\sum_{n=0}^k 10^n)_{10} $$
The magic trick here, but where my main mistake may be, is that:
$$ x_{10} > y_{10} \iff ((\beta-1)\sum_{n=0}^k \beta^n)_{10} > ((\beta-1)\sum_{n=0}^k 10^n)_{10} $$
And we finally get:
$$ x_{10} > y_{10} \\ \iff \\ (\sum_{n=0}^k \beta^n)_{10} > (\sum_{n=0}^k 10^n)_{10} \\ \iff \\ \beta > 10 $$
If this process is repeated I find that this iterative function is not bijective, which basically makes the whole reduction thing a loss of time. Why is this just injective?
Thanks.