I always liked this proof, from the theory of Umbral calculus developed by Rota (See "Combinatorics: The Rota Way", by Joseph Kung, Gian Carlo Rota and Catherin Yan, chapter 4.2)

**Proposition**: Let $(a_n)_{n\geq 0}$ and $(b_n)_{n \geq 0}$ be sequences. Then
$$b_n=\sum_{k=0}^n\binom{n}{k} a_k  \ \text{ for all } n \Longleftrightarrow a_n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b_k \ \text{ for all } n.$$

The heuristic proof use the notion of "raising and lowering subscripts and superscript". Raising subscripts at the right side we obtain
$$b^n=\sum_{k=0}^n\binom{n}{k}a^k=(a+1)^n.$$
Hence, for all $n$,
$$a^n=(b-1)^n=\sum_{k=0}^n (-1)^{n-k}\binom{n}{k}b^k.$$
Lowering exponents, we obtain the inverse relation.