I noticed that for the binomial theorem, you can in some sense "swap" the locations of the index expressions, like so
$$(x+y)^n = \sum_{k=0}^{n}{n \choose k}x^ky^{n-k} = \sum_{k=0}^{n}{n \choose k}x^{n-k}y^{k}$$ where here we see the $k$ and $n-k$ in the exponents can be exchanged yet maintain the equivalence.
Clicking through random wiki links, I then stumbled across a recurrence relation for complete bell polynomials, $$B_{n+1}(x_1,x_2,...,x_{n+1})= \sum_{i=0}^{n}{n \choose i}B_{n-i}(x_1,x_2,...,x_{n-i})x_{i+1}$$
and wonder, similar to the binomial theorem, if one can exchange the index expressions to equivalently yield
$$B_{n+1}(x_1,x_2,...,x_{n+1})= \sum_{i=0}^{n}{n \choose i}B_{i+1}(x_1,x_2,...,x_{i+1})x_{n-i}$$ where the $i+1$ and $n-i$ terms exchange places.
If this is true, how do we know and to what extent can this index exchange property be generalized?
Based on: $$\binom{n}i=\frac{n!}{i!(n-i)!}=\binom{n}{n-i}$$we find: $$\sum_{i=0}^{n}{n \choose i}B_{n-i}(x_{1},x_{2},...,x_{n-i})x_{i+1}=\sum_{i=0}^{n}{n \choose n-i}B_{n-i}(x_{1},x_{2},...,x_{n-i})x_{i+1}=$$$$\sum_{j=0}^{n}{n \choose j}B_{j}(x_{1},x_{2},...,x_{j})x_{n-j+1}$$where in the second equality we substituted $j$ for $n-i$.
Of course we can replace $j$ for $i$ in the last expression to end up with:$$\sum_{i=0}^{n}{n \choose i}B_{n-i}(x_{1},x_{2},...,x_{n-i})x_{i+1}=\sum_{i=0}^{n}{n \choose i}B_{i}(x_{1},x_{2},...,x_{i})x_{n-i+1}$$
It comes to the same as replacing $i$ by $n-i$ everywhere in the expression except for coefficient $\binom{n}i$ (where a replacement is somehow redundant because it does not change the value).