Hence conjecture, and prove by mathematical induction, that dⁿ y / dxⁿ = (-1)ⁿ (n-1)! / xⁿ for all positive integers n.

We need to show that the formula holds for $n=1$.
The formula states:

$$\frac{d^1 y}{dx^1} = (-1)^1 \frac{(1-1)!}{x^1} = (-1) \frac{0!}{x} = -\frac{1}{x}$$

Now, let's find the first derivative of $y = -\ln x$:

$$\frac{dy}{dx} = \frac{d}{dx}(-\ln x) = -\frac{1}{x}$$

Since both sides are equal, the formula holds for $n=1$.

Assume that the formula holds for some positive integer $k$. That is, assume:

$$\frac{d^k y}{dx^k} = (-1)^k \frac{(k-1)!}{x^k}$$

We need to show that if the formula holds for $k$, it also holds for $k+1$.
We want to show that:

$$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^{k+1} \frac{((k+1)-1)!}{x^{k+1}} = (-1)^{k+1} \frac{k!}{x^{k+1}}$$

We know that the $(k+1)$-th derivative is the derivative of the $k$-th derivative:

$$\frac{d^{k+1} y}{dx^{k+1}} = \frac{d}{dx} \left( \frac{d^k y}{dx^k} \right)$$

Using our inductive hypothesis, substitute the expression for $\frac{d^k y}{dx^k}$:

$$\frac{d^{k+1} y}{dx^{k+1}} = \frac{d}{dx} \left( (-1)^k \frac{(k-1)!}{x^k} \right)$$

Rewrite $\frac{1}{x^k}$ as $x^{-k}$:

$$\frac{d^{k+1} y}{dx^{k+1}} = \frac{d}{dx} \left( (-1)^k (k-1)! x^{-k} \right)$$

Since $(-1)^k (k-1)!$ are constants with respect to $x$, we can pull them out of the derivative:

$$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^k (k-1)! \frac{d}{dx} (x^{-k})$$

Now, differentiate $x^{-k}$ with respect to $x$:

$$\frac{d}{dx} (x^{-k}) = -k x^{-k-1}$$

Substitute this back into the equation:

$$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^k (k-1)! (-k x^{-k-1})$$

Rearrange the terms:

$$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^k (-1) k (k-1)! x^{-(k+1)}$$ $$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^{k+1} k (k-1)! x^{-(k+1)}$$

Since $k (k-1)! = k!$, we can write:

$$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^{k+1} k! x^{-(k+1)}$$ $$\frac{d^{k+1} y}{dx^{k+1}} = (-1)^{k+1} \frac{k!}{x^{k+1}}$$

This is the desired form for $n=k+1$. Thus, if the formula holds for $k$, it also holds for $k+1$.

By the principle of mathematical induction, the statement $\frac{d^n y}{dx^n} = (-1)^n \frac{(n-1)!}{x^n}$ is true for all positive integers $n$, given $y = -\ln x$.

The final answer is $\boxed{\text{Proof by mathematical induction completed.}}$