Consider the Banach algebra $\ell^{1}(\mathbb{Z})$ with multiplication $$(x\ast y)_{k}:=\sum_{l\in\mathbb{Z}}x_{k-l}y_{l}.$$ This may seem strange, but I know how to solve the following problem. I'm just looking for a neat short argument because I think that should be possible. Here is the problem:
If $e_{n}$ denotes the standard $n$-th coordinate sequence, then $e_{n}=e_{1}^{n}$ for all $n\in\mathbb{Z}$.
I only know a way to prove it by distinguishing some cases and it bothers me because I think the argument can be really short. So I was wondering if there was a more straightforward way (maybe a one-line proof).
Thanks in advance.
Hint: Show that $e_n*e_m=e_{n+m}$. The result then follows.