Consider two operators A and B.
Assume [A,B] = C,  [A,C] = [B,C] = 0.
Often C is just a number times the identity operator.

[A, exp(gB)] = [A, 1 + gB + (g²/2!)B² + (g³/3!)B³ + ... ]
= gC + (g²/2!)2BC + (g³/3!)3B²C + ...
= gCexp(gB).

[A,B²] = CB-BC = 2BC.
[A,Bⁿ] = nCB^n-1 holds for n = 1 and 2.
We show that if it holds for n it also holds for n+1.
Then [A,Bⁿ⁺¹] = B[[A,Bⁿ] + [A,B]Bⁿ = BnCB^n-1 + CBⁿ = (n+1)CBⁿ.
Therefore it holds for all n.

Let f(x) = e^Axe^Bx,  with x a continuous variable.
df/dx = Ae^Axe^Bx + e^Axe^BxB = f(x)(e^-BxAe^Bx + B)
= f(x) (e^-Bx(e^BxA + x[A,B]e^Bx + B)) = f(x)(A + x[A,B] + B)
using the expression we proved above.

df/dx = f(x)(A + x[A,B] + B) is a differential equation with solution
f(x) = e^(A+B)xexp(½x²[A,B]).
For x = 1 we therefore have e^{(A+B) =} e^Ae^B exp(-½x²[A,B]).