I was having a look at the paper of Maynard https://arxiv.org/pdf/1604.01041.pdf where he proves that there are infinite primes without a certain digit $a_0$. The paper starts with the result that, given $a_0\in\{0,\dots,9\}$, then $$ {\mathcal{A_1(x)}=\#\left\{ m=\sum_{0\leq i\leq k} n_i 10^i:\; n_i\in\{0,\dots,9\}\setminus\{a_0\}\,,\;k\geq0\,,\;m\leq x \right\}}=O(x^{1-c}) $$ where $$ c=\frac{\log(10/9)}{\log 10} \thickapprox 0.046\;. $$ This seems to be a well-known result, maybe even elementary, but I wasn't able to figure/find out any proof of it. Any suggestion?
Thanks in advance
If you had $m\lt x$ instead of $m\le x$, then for $x=10^k$ the set would have exactly $x^{1-c}$ elements: There are $9$ choices for each digit, so the set has $9^k=9^{\log x/\log10}=x^{\log9/\log10}=x^{1-c}$ elements. Since the number of elements in the set grows monotonically, its size is bounded by $10^{(k+1)(1-c)}$ for $x\in\left[10^k,10^{k+1}\right]$, and thus by $10^{1-c}x^{1-c}=O\left(x^{1-c}\right)$.