Step 1: Frame it as repeated trials.
A die is rolled four times; success means a perfect square. This is a binomial situation with $n=4$.
Step 2: Find the success probability.
Perfect squares on a die are $1$ and $4$, so $p = \dfrac{2}{6} = \dfrac{1}{3}$ and failure $q = \dfrac{2}{3}$.
Step 3: Use the complement.
At least one success is easiest as $1 - P(\text{no success})$, where no success means $X = 0$.
Step 4: Apply the binomial term for $X=0$.
$P(X=0) = \binom{4}{0}p^0 q^4 = q^4 = \left(\dfrac{2}{3}\right)^4 = \dfrac{16}{81}$.
Step 5: Subtract from one.
$P(X\ge 1) = 1 - \dfrac{16}{81} = \dfrac{65}{81}$.
Step 6: Reflect.
Using the binomial $X=0$ term is the same as multiplying the per-throw failure probability four times, confirming the answer.
\[ \boxed{\dfrac{65}{81}} \]