The following theorem [S. L. Ross] does not require $f(n)$ be of exponential order. Does it have some explanation?
Let $f$ be a real valued function having a continuous $(n-1)$st derivative $f^{(n-1)}$ for $t\geq 0$; and assume that $f,\;f',\;f'',\ldots, f^{(n-1)}$ are all of exponential order $e^{\alpha t}$. Suppose $f^{(n)}$ is piecewise continuous in every finite closed interval $0\leq t \leq b$. Then $\mathcal{L} \{f^{(n)}\}$ exists for $s>\alpha$ and $$\mathcal{L} \{f^{(n)}\} =s^n\mathcal{L} \{f\}-s^{n-1}f(0)-s^{n-2}f'(0)-s^{n-3} f''(0)-\ldots -f^{(n-1)}(0)$$
A function Laplace transform exists if it is piecewise continuous and of exponential order. Here we are interested in Laplace transform of $f^{(n)}$, so it must be piecewise continuous and must be of exponential order. The problem here is that the theorem does not require exponential order property for $f^{(n)}$ even though it requires this to be satisfied for lower order derivatives of $f$.