I know as $n \to \infty$, this probability is $0$, I know this because there are approximately $\log_{10}(n)$ powers of $10$ less than $n$ and so the probability of a natural number in the interval $[1, n]$ being a power of $10$ is approx $\frac{\log_1(n)}{n}$ which goes to zero as $n \to \infty$. Thus the answer is $0$.
I was wondering if there was a simpler way of doing this? Rather than using logs? Maybe sum to infinity of a geometric progression? Could anyone help!? Thank you!
You may answer this question by considering a "uniform" distribution over natural numbers that is a charge (finitely additive probability measure) on $(\mathbb{N},2^{\mathbb{N}})$. Specifically, for $A\subset \mathbb{N}$ let $(\delta_A)_n=\mathbf{1}_A(n)$ and let $\mathsf{P}(A):=T(\delta_A)$, where $T$ is a Banach limit on $\ell^{\infty}$.
For example, $\mathsf{P}(\mathbb{N})=1$, $\mathsf{P}(A)=0$ for any finite set $A$, and $\mathsf{P}(\{n\in \mathbb{N}:n\equiv1 \mod 2\})=1/2$.
Now, using your argument, $\mathsf{P}(\{n\in\mathbb{N}:n\text{ is a power of $10$}\})=0$.