[proofplan] We prove the base case $n = 1$ by integration by parts, using the exponential order hypothesis to kill the boundary term at infinity. The general formula follows by induction, applying the base case to $f^{(n-1)}$ and substituting the inductive hypothesis. [/proofplan] [step:Establish the base case by integration by parts] Write out the definition of $\mathcal{L}\{f'\}(p)$ and integrate by parts with $u = e^{-pt}$ and $dv = f'(t)\,d\mathcal{L}^1(t)$: \begin{align*} \mathcal{L}\{f'\}(p) = \int_0^\infty f'(t)e^{-pt}\,d\mathcal{L}^1(t) = \bigl[f(t)e^{-pt}\bigr]_0^\infty + p\int_0^\infty f(t)e^{-pt}\,d\mathcal{L}^1(t). \end{align*} Since $f$ is of exponential order, there exist $M > 0$ and $\sigma_0 \in \mathbb{R}$ with $|f(t)| \leq Me^{\sigma_0 t}$ for all $t \geq 0$. For $\operatorname{Re} p > \sigma_0$: \begin{align*} |f(t)e^{-pt}| \leq Me^{\sigma_0 t}e^{-(\operatorname{Re} p)t} = Me^{(\sigma_0 - \operatorname{Re} p)t} \to 0 \quad \text{as } t \to \infty. \end{align*} Hence the boundary term at $t = \infty$ vanishes. At $t = 0$, the boundary term is $f(0)e^0 = f(0)$. Therefore: \begin{align*} \mathcal{L}\{f'\}(p) = -f(0) + p\hat{f}(p) = p\hat{f}(p) - f(0). \end{align*} [guided] The idea is to move the derivative off $f'$ and onto the exponential kernel $e^{-pt}$. Integration by parts with $u = e^{-pt}$ and $dv = f'(t)\,d\mathcal{L}^1(t)$ gives $du = -pe^{-pt}\,d\mathcal{L}^1(t)$ and $v = f(t)$, so \begin{align*} \int_0^\infty f'(t)e^{-pt}\,d\mathcal{L}^1(t) = \bigl[f(t)e^{-pt}\bigr]_0^\infty - \int_0^\infty f(t)(-p)e^{-pt}\,d\mathcal{L}^1(t) = \bigl[f(t)e^{-pt}\bigr]_0^\infty + p\int_0^\infty f(t)e^{-pt}\,d\mathcal{L}^1(t). \end{align*} The second integral is exactly $p\hat{f}(p)$. We must verify the boundary term vanishes at $t = \infty$. The exponential order hypothesis gives $|f(t)| \leq Me^{\sigma_0 t}$, so for $\operatorname{Re} p > \sigma_0$: \begin{align*} |f(t)e^{-pt}| \leq Me^{(\sigma_0 - \operatorname{Re} p)t} \to 0 \quad \text{as } t \to \infty, \end{align*} since $\sigma_0 - \operatorname{Re} p < 0$. At $t = 0$: $f(0)e^{0} = f(0)$. Combining: \begin{align*} \mathcal{L}\{f'\}(p) = 0 - f(0) + p\hat{f}(p) = p\hat{f}(p) - f(0). \end{align*} This is the key identity: differentiation in the time domain becomes multiplication by $p$ in the transform domain, with the initial value $f(0)$ appearing as an additive correction. [/guided] [/step] [step:Iterate to obtain the general formula by induction on $n$] Suppose the formula holds for $n - 1$: \begin{align*} \mathcal{L}\{f^{(n-1)}\}(p) = p^{n-1}\hat{f}(p) - \sum_{k=0}^{n-2} p^{n-2-k}f^{(k)}(0). \end{align*} Apply the base case to the function $g = f^{(n-1)}$, which is continuously differentiable with $g' = f^{(n)}$ of exponential order: \begin{align*} \mathcal{L}\{f^{(n)}\}(p) = \mathcal{L}\{g'\}(p) = p\mathcal{L}\{g\}(p) - g(0) = p\,\mathcal{L}\{f^{(n-1)}\}(p) - f^{(n-1)}(0). \end{align*} Substituting the inductive hypothesis: \begin{align*} \mathcal{L}\{f^{(n)}\}(p) &= p\left(p^{n-1}\hat{f}(p) - \sum_{k=0}^{n-2} p^{n-2-k}f^{(k)}(0)\right) - f^{(n-1)}(0) \\ &= p^n\hat{f}(p) - \sum_{k=0}^{n-2} p^{n-1-k}f^{(k)}(0) - f^{(n-1)}(0) \\ &= p^n\hat{f}(p) - \sum_{k=0}^{n-1} p^{n-1-k}f^{(k)}(0). \end{align*} This completes the induction. [/step]