First, find the first derivative:

\(\frac{d}{dx}(x \ln x) = 1 + \ln x\).

Check the base case for \(n = 2\):

\(\frac{d^2}{dx^2}(x \ln x) = x^{-1} = (-1)^2 (2-2)! x^{1-2}\).

Assume that \(\frac{d^k}{dx^k}(x \ln x) = (-1)^k (k-2)! x^{1-k}\) holds for some integer \(k \geq 2\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1}}{dx^{k+1}}(x \ln x) = (-1)^k (k-2)!(1-k)x^{1-k-1}\).

Simplify:

\(= (-1)^{k+1} ((k+1)-2)! x^{1-(k+1)}\).

Thus, \(\frac{d^n}{dx^n}(x \ln x) = (-1)^n (n-2)! x^{1-n}\) is also true for \(n = k+1\).

By induction, the statement is true for every integer \(n \geq 2\).