It it easy to prove $$x^n=\sum_{i=0}^{n}S(n,i)x_{(i)}$$ by induction or recursive formula...
But is there some proof without word for it? or at least a direct proof that concern with partitions of sets? or is there any intuition behind it?
2026-02-24 05:31:02.1771911062
Proof without words for Stirling formula
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Yes, there is an intuition. You know that $x_{(i)}=\frac{x!}{(x-i)!},$ if you multiply and divide by $i!$ you get this expression $$x^n=\sum_{i=0}^{n}S(n,i)i!\frac{x!}{(x-i)!i!}=\sum_{i=0}^{n}S(n,i)i!\binom{x}{i},$$ You can think in why $S(n,i)i!$ represents surjective functions from $[n]=\{1,2,\cdots ,n\}$ to $[i]=\{1,2,\cdots ,n\}.$
So, basically, you are taking functions $f:[n]\longrightarrow[x]$ which are in the set $[x]^{[n]}$(s.t $|[x]^{[n]}|=x^n$) and seeing what's the cardinality of the image. If the cardinality is $i$, those functions can be represented in $S(n,i)i!\binom{x}{i}$.