'any sequence of $n$ partial differentiations of $f$ results in a constant times $f^{n}$'

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Ch5.1

In the last part of Ch5.1 of the text, just before a corollary (Cor 5.5), it says:

'$\color{red}{\text{any sequence of $n$ partial differentiations of $f$ results in a constant times $f^{n}$}}$'

Question: What is the constant, and how do we prove it?

Here is what I tried.

$n=1$

$$f'=f_x=-if_y \implies f_x = f', f_y=if'$$

$n=2$

$$f''=(f_x)'=f_{xx}$$

$$f''=(-if_y)'=-i(f_{xy})=-i(f_{yx})$$

$$f''=(-i)^2(f_y)''$$

I guessed the last one but deduced the others from definition of derivative. I also guess

$$f^{(n)} = (-i)^{n-k}1^kf_{x^{(k)}y^{(n-k)}} \implies \frac{1}{(-i)^{n-k}1^k}f^{(n)} = f_{x^{(k)}y^{(n-k)}}$$

$$\therefore, \ \text{the constant is} \ \frac{1}{(-i)^{n-k}1^k}.$$

Not sure if there's supposed to be a $\binom{n}{k}$.

As for the proof, I guess by induction with definition of derivatives: something like if it holds true for $k, n-k$, then we must show it's true for $k+1,n-k$ ($\frac{\partial}{\partial x}$) and $k,n-(k-1)$ ($\frac{\partial}{\partial y}$).

Answer 1

Let $s = (s_1,\ldots,s_n)$ by any sequence with $s_i \in \{x, y \}$. Define recursively

$D_{(s_1)}(f) = f_{s_1}, D_{(s_1,\ldots,s_{n+1})}(f) = D_{s_{n+1}}(D_{(s_1,\ldots,s_n)}(f))$.

Then $D_s(f) = i^{y(s)}f^{(n)}$, where $y(s) =$ number of indices $i$ such that $s_i = y$. This is proved by induction.

$n = 1$ : This is well-known.

$n \to n+1$ : $D_{(s_1,\ldots,s_{n+1})}(f) = D_{s_{n+1}}(D_{(s_1,\ldots,s_n)}(f)) = D_{s_{n+1}}(i^{y((s_1,\ldots,s_n))}f^{(n)}) = i^{y((s_1,\ldots,s_n))}D_{s_{n+1}}(f^{(n)}) = i^{y((s_1,\ldots,s_n))}i^{y((s_{n+1}))}f^{(n+1)} = i^{y((s_1,\ldots,s_{n+1})}f^{(n+1)}$.