1. The problem is to solve the equation $n^2 = 7^n$ for $n$ using the Lambert W function. 2. Rewrite the equation: $n^2 = 7^n$ implies $n^2 = e^{n \ln 7}$. 3. Rearrange to isolate terms: $n^2 = e^{n \ln 7}$ can be written as $e^{-n \ln 7} n^2 = 1$. 4. Define $x = n \ln 7$ so that $n = \frac{x}{\ln 7}$. Substitute to get: $$e^{-x} \left(\frac{x}{\ln 7}\right)^2 = 1$$ 5. Multiply both sides by $e^{x}$ to get: $$\left(\frac{x}{\ln 7}\right)^2 = e^{x}$$ 6. Rewrite as: $$x^2 = (\ln 7)^2 e^{x}$$ 7. Rearranged to isolate $x$: $$\frac{x^2}{e^{x}} = (\ln 7)^2$$ 8. Multiply both sides by $e^{-x}$: $$x^2 e^{-x} = (\ln 7)^2$$ 9. Set $y = -x$, so: $$x^2 e^{-x} = (-y)^2 e^{y} = y^2 e^{y} = (\ln 7)^2$$ 10. Now focus on solving: $$y^2 e^{y} = (\ln 7)^2$$ 11. Make substitution $z = y e^{y/2}$. Then $z^2 = y^2 e^{y} = (\ln 7)^2$, so $z = \pm \ln 7$. 12. From $z = y e^{y/2} = \pm \ln 7$, express as: $$y e^{y/2} = \pm \ln 7$$ 13. Let $w = \frac{y}{2}$, then: $$2 w e^{w} = \pm \ln 7$$ 14. So: $$w e^{w} = \pm \frac{\ln 7}{2}$$ 15. By definition of Lambert W function, $w = W_k\left( \pm \frac{\ln 7}{2} \right)$ where $k$ denotes the branch. 16. Recall $w = \frac{y}{2} = \frac{-x}{2}$, so: $$x = -2 W_k\left( \pm \frac{\ln 7}{2} \right)$$ 17. Finally, recall $n = \frac{x}{\ln 7}$, hence: $$n = \frac{-2}{\ln 7} W_k\left( \pm \frac{\ln 7}{2} \right)$$ 18. Therefore, the solutions for $n$ are: $$n = \frac{-2}{\ln 7} W_k\left( \frac{\ln 7}{2} \right) \quad \text{and} \quad n = \frac{-2}{\ln 7} W_k\left( -\frac{\ln 7}{2} \right)$$ where $W_k$ is the Lambert W function on the appropriate branch(s). These values give the solutions of the original equation $n^2 = 7^n$ using the Lambert W function.