# Combinatorial proof for an identity about Stirling cyclic numbers

Solving through Lovász' Combinatorial Problems and Exercises I found an exercise asking me to prove two identities:

$$\sum_{k = 0}^n {n \brace k} (x)_n = x^n$$ $$\sum_{k = 0}^n \left[ n \atop k \right] x^k = x^{(n)}$$

(Notation note: I'm using Pochhamer symbols, brackety ones are cyclic Stirling numbers and curly ones are partition Stirling numbers)

First one yields itself to a combinatorial proof, basically a combinatorial interpretation of the fact that every function can be ''transformed'' into a surjection if we restrict its codomain.

I was able to prove second one using generating functions, but I was unable to find a combinatorial proof. Due to obvious analogies between two identities (writing powers like a sum of falling factorials/ writing rising factorials like a sum of powers) I'm curious is there a combinatorial proof to the second identity?