Let $f : \mathbb{R} \to \mathbb{C}$ be a function of normalized bounded variation (NBV), meaning that $f$ is of bounded variation, $f$ is right continuous, and $f(x) \to 0$ and $x \to -\infty$. As explained in Section 3.5 of Folland's Real Analysis textbook, there is a unique complex measure $df$ with the property that $df(x_1, x_2] = f(x_2) - f(x_1)$ for all $x_1 < x_2$ in $\mathbb{R}$.
Moreover, in Exercise 34, which accompanies this section, we are asked to prove that for any two NBV functions $f$ and $g$, we have $$d(fg) = \tfrac{1}{2}(g(x+) + g(x-))df + \tfrac{1}{2}(f(x+) + f(x-))dg,$$ where $f(x\pm)$ denotes the right and left hand limits of $f$, respectively.
I would like to know whether the "power rule" holds for NBV functions. Namely, given NBV $f$ and $n \in \mathbb{N}$, do we have $d(f^n) = \tfrac{n}{2}(f^{n-1}(x+) + f^{n-1}(x-))df$?
It seems plausible that a nice formula like this holds for powers of an NBV function. After all, $f^n$ has the same discontinuities as $f$. The identity is easy to prove for $n = 2$, but I'm not sure it's true for higher $n$. Take $n = 3$ for example. Using the exercise from Folland: $$d(f^3) = \tfrac{1}{2}(f^2(x+) + f^2(x-)) df + \tfrac{1}{2}(f(x+) + f(x-))^2 df.$$
So the difficulty seems that taking powers of a function does not commute nicely with this "symmetric averaging" operation. Is there any way to recover the nicer formula I propose? If not, what consequence does this have for proving a chain rule for something like $e^f$, for real-valued NBV $f$? Is it still possible to show something like $d(e^f) = \tfrac{1}{2}(e^{f(x+)} + e^{f(x-)}) df$? (If I knew the power rule worked I would try to prove this using power series.)
Hints or solutions are greatly appreciated.