Clear["Global`*"]
Defining the operator recursively,
dOp[func_, x_Symbol, 1] := dOp[func, x, 1] = D[func, x] - func;
dOp[func_, x_Symbol, n_Integer?Positive] := dOp[func, x, n] =
D[dOp[func, x, n - 1], x] - n*dOp[func, x, n - 1];
For example,
dOp[f[x], x, 2] // Expand
(* 2 f[x] - 3 f'[x] + f''[x] *)
Looking at the first several,
Table[{n, dOp[f[x], x, n] // Expand}, {n, 1, 6}] //
Grid[#, Alignment -> Left, Dividers -> All] &

The coefficients are Stirling numbers of the first kind, StirlingS1
Table[StirlingS1[n, m], {n, 2, 7}, {m, 1, n}] // Grid

Consequently, the operator can alternatively be written as the sum
dOp2[func_, x_Symbol, n_Integer?Positive] :=
Sum[StirlingS1[n + 1, m + 1] D[func, {x, m}], {m, 0, n}]
Verifying the equivalence of the definitions,
And @@ Table[dOp[f[x], x, n] == dOp2[f[x], x, n] // Simplify, {n, 1, 15}]
(* True *)
D[f,{x,j}]
which takes the $j$th derivative of f w.r.t x, but (for univariate f) you could also meanop[f_, x_, n_] := Product[D[f[x], x] - j, {j, 1, n}]
$\endgroup$