1. **State the problem:** Evaluate the integral $$\int_0^\infty e^{-n} n^5 \, dn$$. 2. **Recall the formula:** This integral is a form of the Gamma function $$\Gamma(k) = \int_0^\infty x^{k-1} e^{-x} \, dx$$ for $$k > 0$$. 3. **Match the integral to the Gamma function:** Here, $$k-1 = 5$$ so $$k = 6$$. 4. **Use the Gamma function property:** For positive integers, $$\Gamma(k) = (k-1)!$$. 5. **Calculate the factorial:** $$\Gamma(6) = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$. 6. **Final answer:** $$\int_0^\infty e^{-n} n^5 \, dn = 120$$.