1. **State the problem:** Find the $n$th derivative of the function $$f(x) = \frac{1}{(x-1)(x-2)(x-3)}.$$\n\n2. **Rewrite the function:** We have $$f(x) = \frac{1}{(x-1)(x-2)(x-3)} = \frac{1}{(x-1)(x-2)(x-3)}.$$\n\n3. **Partial fraction decomposition:** Express $f(x)$ as a sum of simpler fractions:\n$$f(x) = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}.$$\nMultiply both sides by $(x-1)(x-2)(x-3)$ to get:\n$$1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2).$$\n\n4. **Find coefficients $A$, $B$, and $C$ by plugging in $x=1,2,3$: **\n- For $x=1$: $$1 = A(1-2)(1-3) = A(-1)(-2) = 2A \implies A = \frac{1}{2}.$$\n- For $x=2$: $$1 = B(2-1)(2-3) = B(1)(-1) = -B \implies B = -1.$$\n- For $x=3$: $$1 = C(3-1)(3-2) = C(2)(1) = 2C \implies C = \frac{1}{2}.$$\n\n5. **Rewrite $f(x)$ with coefficients:**\n$$f(x) = \frac{1/2}{x-1} - \frac{1}{x-2} + \frac{1/2}{x-3}.$$\n\n6. **Find the $n$th derivative:** The $n$th derivative of $\frac{1}{x-a}$ is given by\n$$\frac{d^n}{dx^n} \left( \frac{1}{x-a} \right) = (-1)^n \frac{n!}{(x-a)^{n+1}}.$$\n\n7. **Apply this to each term:**\n$$f^{(n)}(x) = \frac{1}{2} (-1)^n \frac{n!}{(x-1)^{n+1}} - (-1)^n \frac{n!}{(x-2)^{n+1}} + \frac{1}{2} (-1)^n \frac{n!}{(x-3)^{n+1}}.$$\n\n8. **Factor out common terms:**\n$$f^{(n)}(x) = (-1)^n n! \left( \frac{1}{2(x-1)^{n+1}} - \frac{1}{(x-2)^{n+1}} + \frac{1}{2(x-3)^{n+1}} \right).$$\n\n**Final answer:**\n$$\boxed{f^{(n)}(x) = (-1)^n n! \left( \frac{1}{2(x-1)^{n+1}} - \frac{1}{(x-2)^{n+1}} + \frac{1}{2(x-3)^{n+1}} \right)}.$$