Must there exist a number of digits $n$, for which the first $n$ digits of a normal number repeat?

Question

I understand that if some number $x$ is a normal number, its digits follow a uniform distribution in any base. Additionally, I understand that any particular sequence of $k$ digits of $x$ in base $b$ will occur with density $b^{-k}$

I now wonder whether there must exist some number $n$ for which the sequence digits of $x$ indexed by 1 through $n$ is identical to that indexed by $n+1$ through $2n$.

By this definition, the likelihood of finding the first $n$ digits repeated beginning at digit $n+1$ is $b^{-n}$. Then, the likelihood of there existing such a number $n$ in any of the infinitely many digits of $x$ is

$$P(b) = 1-\prod_{n=1}^\infty (1-b^{-n}).$$

This can alternately be expressed as

$$P(b) = 1 - \left(\frac{1}{b}; \frac{1}{b}\right)_\infty,$$

where $(a;q)_N$ is the q-Pochhammer symbol . I can approximate this expression numerically without much trouble, and the results for a few values of $b$ are below.

$$ \begin{array}{c|c} b & P(b) \\ \hline 2 & 0.71121 \\ \hline 5 & 0.23967 \\ \hline 10 & 0.10999 \\ \hline 16 & 0.066405 \\ \end{array}$$

Clearly, none of these numbers are one. If you're just here for the answer, it's no (unless my reasoning is flawed, and it might be). In bases other than two, most normal numbers do not repeat in this way.

My question: why? What intuition or other manipulation could I apply here? Doing it numerically is pretty inelegant.

In base ten, this reduces to proving that the product $$ (0.9)(0.99)(0.999)... > 0. $$ (Well, I suppose the above is true of any base if we let 9 be the symbol for the largest valued digit.) This is a pretty simple looking product, but I'm a little stuck just demonstrating that it's nonzero. I can rather abstractly believe that the terms in this product approach one faster than the product approaches zero, but that's really hand wavey. The Laurent polynomial expansion is pretty irregular. Maybe I'm missing something obvious; any thoughts would be appreciated. Bonus points if you can show whether $P(b)$ is less than or greater than one half given some $b$.

Answer 1

$$\prod_{i\ge 1} \left(1-10^{-i}\right)=1-\sum_{i\ge 1}10^{-i}+\sum_{i>j\ge 1}10^{-i-j}-\cdots ~~~~(\star)$$

Now, note that $\sum_{i\ge 1}10^{-i} = \frac{1}{10-1}=\frac{1}{9}<1$.

Hence, $\sum_{i>j\ge 1}10^{-i-j} \le \sum_{j\ge 1}10^{-i}\sum_{i\ge 1}10^{-i}=\frac{1}{9}\sum_{i\ge 1}10^{-i} $. Similarly, each summand in $(\star)$ is smaller in absolute value than the previous term. Hence we have an alternating series with terms monotonically approaching $0$.

In particular, that means $$1>\prod_{i\ge 1} \left(1-10^{-i}\right)>1-\sum_{i\ge 1}10^{-i}=1-\frac{1}{9}=\frac{8}{9}$$

All the above easily generalizes if the base isn't $10$.