Answer
See explanation
Work Step by Step
Below is a brief inductive proof that the sequence defined by
\[
a_k \;=\; k\,a_{k-1}
\quad \text{for all integers } k \ge 1,
\quad \text{with} \quad a_0 = 1,
\]
is given explicitly by
\[
\boxed{a_k = k!}.
\]
---
## **Base Case** \((k = 0)\)
We must check that the formula \(a_k = k!\) holds when \(k=0\). By definition,
\[
a_0 = 1,
\]
and by convention in mathematics,
\[
0! = 1.
\]
Hence \(a_0 = 0!\), so the base case is satisfied.
---
## **Inductive Step**
Assume for some integer \(k \ge 1\) that the formula holds at \(k-1\), i.e.,
\[
a_{k-1} = (k-1)!.
\]
We must show that it then holds at \(k\). Using the recursive definition,
\[
a_k
= k \, a_{k-1}.
\]
By the inductive hypothesis, \(a_{k-1} = (k-1)!\). Therefore,
\[
a_k
= k \cdot (k-1)!
= k!.
\]
This completes the inductive step.
