In a book titled "Advances in Problem Solving," authored by Sailesh Shirali, Riordan's identity is mentioned which can also be written as $$S=\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1~~~~(1)$$ Here, we prove it by the integral representation of $(j+1)!$ as: $$S=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{(j+1)!}{n^{j+1}}=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{1}{n^{j+1}} \int_{0}^{\infty} x^{j+1} e^{-x} dx$$ $$\implies S=\frac{1}{n^{n}}\int_{0}^{\infty} x ~(n+x)^{n-1} e^{-x} dx=n^{-n}\int_{0}^{\infty} [(x+n)^n-n(x+n)^{n-1}]~ e^{-x} dx$$ $$S=-n^{-n}\left. (x+n)^n~ e^{-x}\right|_{0}^{\infty}=1.$$
It will be interesting to see other approaches for proving (1).