Call the unique base-ten expansion of a finite number $X$ that doesn’t involve insignificant zeroes the series $$X = \sum_{k=a}^{b} 10^k x_k$$ Therefore, in base ten, the $k$th digit of $X$ is $x_k$ (depending on how you enumerate the digits, but this is not important).
Define $\newcommand{\Exp}{\operatorname{Exp}} \Exp$ to be the operator that returns the sequence $$\Exp X = (x_k)_{k=a}^{b}$$ Sequences are also denoted with braces as in $\Exp X = \{x_k\}_{k=a}^b$; the choice is yours.
Then in set notation we say $$0\notin\Exp X$$
For example, $$\Exp 2^{86} = (4,6,2,5,9,1,1,8,1,7,6,2,6,3,3,5,5,4,2,5,2,1,7,3,7,7)$$ The digits are reversed because in any series $a<b$. If you wanted to make the digits appear in the normal order, you would need to flip the limits of series and change $10^k$ to $10^{-k}$.
This tells us that there exists a sequence $(x_k)_{k=a}^b$ that does not contain $0$ that can be used in the series expansion of $X$. We can notate that statement as
$$\exists (x_k)_{k=a}^b \not\ni 0 : X = \sum_{k=a}^b 10^kx_k$$
and I bet that is what you’re really looking for.