can we have $(\partial _x+f(x))^n=\partial _x^n+\binom{n}{1} (\partial _x)^{n-1}f(x)+\cdots+(f(x))^n$?

Question

We know that $(a+b)^n=a^n+\binom{n}{1}a^{n-1}b+\cdots+b^n$ when $a,b$ commutes to each other.

Am I right?

My question is-

Can we expand $(\partial _x+f(x))^np(x)$ or $(\partial _x+f(x))^n$ in the above formula ?

i.e., can we have $(\partial _x+f(x))^n=\partial _x^n+\binom{n}{1} (\partial _x)^{n-1}f(x)+\cdots+(f(x))^n$ ?

Note that if the field is of characteristic $n$ (prime), then we can have equality.

So when $(\partial _x+f(x))^n$ can be expanded ?

Some hints is given here Formal series expansion of differential operator (d/dx + f(x))^n