I think, contrary to Gerry, that the property is likely to happen. Given an order $a_1\cdots a_{n-1}0$ of the digits, the constraints are that $\sum_1^ka_jn^j\equiv0\,(k)$. It is always satisfied for $k=1$ or $k=n-1$.The probability that this order is OK is $1/(n-2)!$, if the remaining constraints are independent. Therefore the number of good orders is, ``expectedly'', $$\frac{(n-1)!}{(n-2)!}=n-1.$$

Edit. I think, contrary to Gerry, that the property is likely to happen, provided that $n$ is even. Obviously, the last digit must be the zero. Given an order $a_1\cdots a_{n-1}0$ of the digits, the constraints are that $\sum_1^ka_jn^j\equiv0\,(k)$. It is always satisfied for $k=1$. For $k=n-1$, it writes $$\begin{pmatrix} n-1 \\ 2 \end{pmatrix}\equiv0\quad(n-1),$$ whence the condition that $n$ be even.  The probability that the order is OK is $\frac12/(n-2)!$, if the remaining constraints are independent. Therefore the number of good orders is, ``expectedly'', $$\frac12\cdot\frac{(n-1)!}{(n-2)!}=\frac{n-1}2.$$