Proof

Let $n\ge 0$ be an integer.
For $n=0$, we have $0!=\left[\begin{array}{c}0\\0\end{array}\right]=1.$
Let $n > 0.$
By definition of factorial $n!,$ it is the number of permutations of $n$ objects.
By definition of permutations, every permutation $\pi_1,\ldots,\pi_n$ of $\{1,2,\ldots, n\}$ defines an arrangement of cycles.
For if we start with $n_0=n$ and look at $n_1=\pi_{n_0}, n_2=\pi_{n_1},\ldots,$ we eventually come at $n_k=n_0.$
Conversely (and reversing the construction), every cycle arrangement defines a permutation.
Since the Stirling numbers of the first kind $\left[\begin{array}{c}n\\r\end{array}\right]$ can be interpreted as the number of ways to arrange $n$ objects into $r$ cycles, it is therefore also the number of permutations that contain exactly $r$ cycles.
Summing $\left[\begin{array}{c}n\\r\end{array}\right]$ over all $r=0,\ldots,n,$ we must get the total number of permutations of $n$ objects, i.e. $$n!=\sum_{r=0}^n \left[\begin{array}{c}n\\r\end{array}\right],\quad\quad(n\ge 0).$$
∎

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Aigner, Martin: "Diskrete Mathematik", vieweg studium, 1993
Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", Addison-Wesley, 1994, 2nd Edition