**Proof plan.** The proof proceeds by induction on $n$. The base case $n = 1$ is the [Product Rule](/theorems/824). The inductive step differentiates the $n$-th Leibniz formula and applies Pascal's identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$. **Step 1: Base case ($n = 1$).** By the [Product Rule](/theorems/824): \begin{align*} (uv)' = u'v + uv' = \binom{1}{0} u' v + \binom{1}{1} u v', \end{align*} which agrees with the formula. **Step 2: Inductive hypothesis.** Assume the formula holds for some $n \geq 1$: \begin{align*} (uv)^{(n)} = \sum_{r=0}^{n} \binom{n}{r} u^{(n-r)} v^{(r)}. \end{align*} **Step 3: Inductive step.** Differentiate both sides using the [Product Rule](/theorems/824): \begin{align*} (uv)^{(n+1)} &= \frac{d}{dx} \sum_{r=0}^{n} \binom{n}{r} u^{(n-r)} v^{(r)} \\ &= \sum_{r=0}^{n} \binom{n}{r} \left[ u^{(n+1-r)} v^{(r)} + u^{(n-r)} v^{(r+1)} \right]. \end{align*} Re-index the second sum by setting $s = r + 1$: \begin{align*} (uv)^{(n+1)} &= \sum_{r=0}^{n} \binom{n}{r} u^{(n+1-r)} v^{(r)} + \sum_{s=1}^{n+1} \binom{n}{s-1} u^{(n+1-s)} v^{(s)}. \end{align*} The $r = 0$ term from the first sum contributes $u^{(n+1)}v$. The $s = n+1$ term from the second sum contributes $u v^{(n+1)}$. For $1 \leq r \leq n$, the coefficients combine via Pascal's identity: \begin{align*} \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}. \end{align*} Therefore: \begin{align*} (uv)^{(n+1)} = \sum_{r=0}^{n+1} \binom{n+1}{r} u^{(n+1-r)} v^{(r)}, \end{align*} completing the induction.