I am trying to prove that a number is normal base b $\iff$ it is simply normal in all bases $b^k$ for every integer $k \geq 1$.
I'm a little confused on this because if for example we take a number that is normal is base 3 how would that be simply normal in base 9 as it would not have any digits greater than 3.
These are my definitions for normal and simply normal:
We say a number $x$ in decimal expansion form base-$b$ is simply normal base-$b$ if $\lim_{n \to \infty} \frac{N_n^b(x;{w})}{n} = \frac{1}{b},$ $\forall w \in \{0,1,2,...b-1\}$.
A number $x$ in decimal expansion form base $b$ is normal base-$b$ if for any arbitrary finite string (or word) $w$ with letters from the alphabet $\{0,1,2,...,b-1\}$ $ lim_{n \to \infty}\frac{N_n^b(x;w)}{n} = \frac{1}{b^{|w|}}$ where $|w|$ denotes the lengh of the word.
If somebody could help me get started on how to prove this if and only if statement that would be great.
Update: I have figured out the forward direction, I am still confused on the reverse.