给出两个来自chatgpt的参考证明方法，显然方法二更符合题意（可以直接看方法二）。

方法一：

Let $y=u^n$, where $n$ is a positive integer.

Expanding $\Delta y = (u + \Delta u)^n - u^n$ using the binomial theorem, we get:

$\Delta y = u^n + \binom{n}{1}u^{n-1}(\Delta u) + \binom{n}{2}u^{n-2}(\Delta u)^2 + \cdots + \binom{n}{n-1}u(\Delta u)^{n-1} + (\Delta u)^n - u^n$

Simplifying, we can see that all the terms containing powers of $(\Delta u)$ greater than or equal to $2$ will have a factor of $\Delta u$, so dividing by $\Delta x$ and taking the limit as $\Delta u$ approaches $0$, those terms will disappear. Thus, we get:

$\lim_{\Delta u \to 0} \frac{\Delta y}{\Delta u} = \lim_{\Delta u \to 0} \left(nu^{n-1} + \binom{n}{1}u^{n-2}(\Delta u) + \cdots + \binom{n}{n-2}(\Delta u)^{n-2}u + (\Delta u)^{n-1}\right)$

$= nu^{n-1}$

Therefore, we have shown that $\frac{d}{dx}(u^n) = nu^{n-1}$ for positive integral exponents $n$.

To extend this result to negative integral exponents, we can use the quotient rule. Let $y = \frac{1}{u^n}$. Then, using the power rule that we just proved, we have:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{u^n}\right) = -\frac{n}{u^{n+1}}$

Multiplying both sides by $-1$, we get:

$\frac{d}{dx}(u^{-n}) = (-n)u^{-n-1}$

Thus, we have shown that the power rule holds for all integer exponents.

方法二：

To prove the power rule for positive integral exponents $n$, we start by writing $y=u^n$, where $u$ is a function of $x$. Let us now apply the binomial theorem to expand $\Delta y=(u+\Delta u)^n - u^n$:

$$\Delta y=(u+\Delta u)^n - u^n = \sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^k $$

where ${n\choose k}=\frac{n!}{k!(n-k)!}$ is the binomial coefficient.

Now, dividing by $\Delta x$ yields:

$$\frac{\Delta y}{\Delta x} = \frac{\sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^k}{\Delta x} $$

Using the limit definition of the derivative, we can let $\Delta x$ approach zero:

$$\frac{dy}{dx} = \lim_{\Delta x\to 0} \frac{\sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^k}{\Delta x} $$

We can simplify the numerator by factoring out $(\Delta u)$:

$$\frac{dy}{dx} = \lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}\sum_{k=0}^{n} {n\choose k}u^{n-k}(\Delta u)^{k-1} $$

As $\Delta x$ approaches zero, $\Delta u/\Delta x$ approaches $du/dx$, and $(\Delta u)^{k-1}$ approaches zero for $k\geq 2$（连续函数的性质）. Therefore, we can simplify the expression to:

$$\frac{dy}{dx} = nu^{n-1}\frac{du}{dx} $$

This is the power rule for positive integral exponents.

To extend this result to negative integral exponents, we can use the quotient rule. Let $n$ be a negative integer and $y=u^{-n}$, then:

$$\frac{dy}{dx} = \frac{d}{dx}\left(u^{-n}\right) = \frac{d}{dx}\left(\frac{1}{u^n}\right) $$

Using the quotient rule:

$$\frac{dy}{dx} = -\frac{nu^{n-1}}{u^{2n}}\frac{du}{dx} = -n\frac{du}{dx}u^{-n-1} $$

Thus, the power rule holds for all integer exponents.