= Solution
Write $\mathcal O=\{x:|x|\leq1\}$. First, $\mathcal O$ is compact. A compact neighborhood of zero contains a sufficiently small closed ball $t^m\mathcal O$, since $|t|^m\to0$; that ball is closed in the compact neighborhood. Scaling back proves compactness of $\mathcal O$. Also $K$ is complete: a <Cauchy sequence> has a tail in a translate of a compact neighborhood, hence a convergent subsequence, and the Cauchy condition makes the whole sequence converge. This is the <completeness of locally compact nontrivially valued fields> argument.
The additive subgroup $t\mathcal O$ is open: a sufficiently small ball around any of its points stays in $t\mathcal O$ by the ultrametric inequality. Thus the compact discrete quotient $\mathcal O/t\mathcal O$ is finite. Choose a set $A\subset\mathcal O$ of representatives, with zero representing the zero coset. For $x\in\mathcal O$, recursively choose $a_n\in A$ and $x_{n+1}\in\mathcal O$ such that $x_n=a_n+t x_{n+1}$, beginning with $x_0=x$. Then
$$
x=\sum_{j=0}^{m-1}a_jt^j+t^m x_m,\qquad |t^m x_m|\leq|t|^m\longrightarrow0.
$$
For arbitrary $x\in K$, multiply first by a large power of $t$ to put it in $\mathcal O$, then divide the resulting expansion by that power. This gives the <digit expansion with a nonuniformizer>.
For uniqueness, take the first exponent $r$ at which two expansions differ. After dividing their difference by $t^r$, all the later terms belong to the closed subgroup $t\mathcal O$, whereas $a_r-b_r$ does not, because the representatives are distinct. The difference cannot be zero. Conversely every series with coefficients in $A$ and only finitely many nonzero negative-index terms converges by completeness and $|t|<1$. \b[Every element therefore has a unique expansion $\boxed{x=\sum_{n\geq r}a_n t^n}$], with the harmless initial zero coefficients understood as part of the same indexed series. No assumption that $t$ is a <uniformizer> is needed.
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