Proving Riordan's identity that $\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1$ 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).

We seek to show that
$$S= \sum_{k=0}^{n-1} {n-1\choose k} \frac{(k+1)!}{n^{k+1}} = 1.$$
The sum is
$$\frac{(n-1)!}{n}
\sum_{k=0}^{n-1} \frac{k+1}{(n-1-k)!} \frac{1}{n^k}.$$
We get without the factor  in front
$$\sum_{k=0}^{n-1} \frac{-n+k+1}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{n}{(n-1-k)!} \frac{1}{n^k}
\\ = - \sum_{k=0}^{n-2} \frac{n-1-k}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = - \sum_{k=0}^{n-2} \frac{1}{(n-2-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = - \sum_{k=1}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
+ \sum_{k=0}^{n-1} \frac{1}{(n-1-k)!} \frac{1}{n^{k-1}}
\\ = \frac{1}{(n-1)!} \frac{1}{n^{-1}}.$$
Restoring the factor in front we find
$$\frac{(n-1)!}{n} \times
\frac{1}{(n-1)!} \frac{1}{n^{-1}} = 1$$
as claimed.
