Here is a simple but cute computation that makes a good exercise in elementary representation theory with a probabilistic flavor — as usual, it’s probably well-known, but I wasn’t aware of it until doing the exercise myself.
We consider random permutations in the symmetric group $latex S_n$, and write $latex \omega_n(\sigma)$ for the number of cycles in the disjoint cycle representation of $latex \sigma\in S_n$. It is a well-known fact (which may be due to Feller) that, viewed as a random variable on $latex S_n$ (with uniform probability measure), there is a decomposition in law
$latex \omega_n=B_1+\cdots+B_n,$
where $latex B_i$ is a Bernoulli random variable taking value $latex 0$ with probability $latex 1-1/i$ and $latex 1$ with probability $latex 1/i$, and moreover the $latex (B_i)$’s are independent.
This decomposition is (in the sources I know) usually proved using the “Chinese Restaurant Process” to describe random permutations ($latex n$ guests numbered from $latex 1$ to $latex n$ enter a restaurant with circular tables; they successively sit either at the next available space of one of the tables already occupied, or pick a new one, with the same probability; after all $latex n$ guests are seated, reading the cycles on the occupied tables gives a uniformly distributed element of $latex S_n$.) If one is only interested in $latex \sigma_n$, then the decomposition above is equivalent to the formula
$latex \mathbf{E}(z^{\omega_n})=\prod_{j=1}^n\Bigl(1-\frac{1}{j}+\frac{z}{j}\Bigr)$
for the probability generating function of $latex \omega_n$. (This formula comes up in mod-poisson convergence of $latex \sigma_n$ and in analogies with the Erdös-Kac Theorem, see this blog post).
Here is an algebraic proof of this generating function identity. First, $latex z\mapsto \mathbf{E}(z^{\omega_n})$ is a polynomial, so it is enough to prove the formula when $latex z=m$ is a non-negative integer. We can do this as follows: let $latex V$ be a vector space of dimension $latex m$ over $latex \mathbf{C}$. Then define $latex W=V^{\otimes n}$ and view it as a representation space of $latex S_n$ where the symmetric group permutes the factors in the tensor product. Using the “obvious” basis of $latex W$, it is elementary that the character of this representation is the function $latex g\mapsto m^{\sigma_n(g)}$ on $latex S_n$ (the matrix representing the action of $latex g$ is a permutation matrix). Hence, by character theory, $latex \mathbf{E}(m^{\sigma_n})$ is the dimension of the $latex S_n$-invariant subspace of $latex W$. Since the representation is semisimple, this is the dimension of the coinvariant space, which is the $latex n$-th symmetric power of $latex V.$ This in turn is well-known (number of monomials of degree $latex n$ in $latex m$ variables), and we get
$latex \mathbf{E}(m^{\omega_n})=\binom{m+n-1}{n}=\prod_{j=1}^n\Bigl(1-\frac{1}{j}+\frac{m}{j}\Bigr),$
as claimed!
