1. **Problem Statement:** Prove that the derivative of $x^n$ with respect to $x$ is $nx^{n-1}$. 2. **Formula Used:** The derivative of a function $f(x)$ is defined as $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition to $f(x) = x^n$:** $$\frac{d}{dx} x^n = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$$ 4. **Expand $(x+h)^n$ using the Binomial Theorem:** $$(x+h)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} h^k = x^n + nx^{n-1}h + \binom{n}{2} x^{n-2} h^2 + \cdots + h^n$$ 5. **Substitute the expansion back:** $$\frac{(x+h)^n - x^n}{h} = \frac{x^n + nx^{n-1}h + \binom{n}{2} x^{n-2} h^2 + \cdots + h^n - x^n}{h}$$ 6. **Simplify numerator:** $$= \frac{nx^{n-1}h + \binom{n}{2} x^{n-2} h^2 + \cdots + h^n}{h}$$ 7. **Divide each term by $h$:** $$= nx^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + h^{n-1}$$ 8. **Take the limit as $h \to 0$:** All terms with $h$ vanish, leaving $$\lim_{h \to 0} \left(nx^{n-1} + \binom{n}{2} x^{n-2} h + \cdots + h^{n-1}\right) = nx^{n-1}$$ 9. **Conclusion:** Therefore, $$\frac{d}{dx} x^n = nx^{n-1}$$ This completes the proof in a simple and clear way.