1. Given the equation $$a^{n^{n}} + bn^{n} + c = 0$$, we want to express $$n$$ in terms of the Lambert W function. 2. First, isolate the term containing $$n$$: $$a^{n^{n}} = -bn^{n} - c$$. 3. It is complex to solve in this form because $$n^{n}$$ appears both in the exponent and as the base's power. 4. Let $$x = n^{n}$$. Then the equation becomes $$a^{x} + bx + c = 0$$, or $$a^{x} = -bx - c$$. 5. Rewrite $$a^{x}$$ as $$e^{x\ln a}$$, so $$e^{x\ln a} = -bx - c$$. 6. Rearrangement yields a transcendental equation involving $$x$$ that is difficult to invert directly. 7. To apply the Lambert W function, we look for an expression of the form $$z = We^{W}$$. 8. Such a direct manipulation is complicated due to the $$-bx - c$$ term on the right side. 9. If we assume $$c = 0$$ for simplification, the equation reduces to $$e^{x\ln a} = -bx$$. 10. Then $$x e^{-x \ln a} = -\frac{1}{b}$$. 11. Multiply both sides by $$-\ln a$$: $$-x \ln a e^{-x \ln a} = \frac{\ln a}{b}$$. 12. Let $$u = -x \ln a$$, then $$u e^{u} = \frac{\ln a}{b}$$. 13. By definition of Lambert W, $$u = W\left(\frac{\ln a}{b}\right)$$. 14. Substituting back, $$-x \ln a = W\left(\frac{\ln a}{b}\right)$$, so $$x = -\frac{1}{\ln a} W\left(\frac{\ln a}{b}\right)$$. 15. Remember $$x = n^{n}$$, so $$n^{n} = -\frac{1}{\ln a} W\left(\frac{\ln a}{b}\right)$$. 16. To solve for $$n$$ in terms of Lambert W remains difficult, but rewriting $$n$$ using Lambert W for $$n^{n} = d$$ is known as: $$n = \frac{\ln d}{W(\ln d)}$$. 17. Thus the final expression for $$n$$ is $$n = \frac{\ln \left(-\frac{1}{\ln a} W\left(\frac{\ln a}{b}\right)\right)}{W\left(\ln \left(-\frac{1}{\ln a} W\left(\frac{\ln a}{b}\right)\right)\right)}$$. **Final answer:** $$n = \frac{\ln \left(-\frac{1}{\ln a} W\left(\frac{\ln a}{b}\right)\right)}{W\left(\ln \left(-\frac{1}{\ln a} W\left(\frac{\ln a}{b}\right)\right)\right)}$$ This expression assumes $$c=0$$ and that all parameters and principal branches are appropriately defined for real values.