Answer
\[
\boxed{a_k = k!}.
\]
Work Step by Step
From the recursion \(a_{k} = k \, a_{k-1}\) with \(a_{0}=1\), you can see by unwrapping (iterating) that
\[
a_k
= k \cdot a_{k-1}
= k \cdot (k-1) \, a_{k-2}
= k \cdot (k-1) \cdot (k-2) \, a_{k-3}
= \cdots
= k \cdot (k-1) \cdot (k-2) \cdots 1 \cdot a_0
= k!.
\]
