A sequence $u=(u_n)_{n\in\mathbb{N}}$ is called exponential if $u_0=1$ and for any integers $m$ and $n$, $u_nu_m=\binom{n+m}{n}u_{n+m}$.
I want to show that the set of exponential sequences is a group with the product $u\cdot v=w$, where $w$ is defined by
$$w_n=\sum_{i+j=n}u_iv_j$$
I'm missing a clean proof of the product of two exponential sequances is itself exponential. The unit is the sequence $(1,0,0,0....)$ and the inverse should be given by $u^{-1}=( (-1)^nu_n)_{n\in\mathbb{N}}$.
Let $f_u(x)$ be the formal power series $$f_u(x) = u_0 + u_1 x + u_2x^2 + \dotsb .$$ Then $f_{u\cdot v}(x) = f_u(x) f_v(x)$ and a sequence $u$ is exponential if and only if $f_u(0)=1$ and $f_u(x)f_u(y) = f_u(x+y)$ for all $x,y$. Use these properties to complete the proof.
[Thanks for pointing out that $f_u(0)=1$ was left out in the original answer.]
Added details: $$f_u(x) f_u(y) = (\sum u_ix^i)(\sum u_j y^j) = \sum_{i,j} u_iu_jx^iy^j= \sum_{n} \sum_{i+j=n} u_iu_jx^iy^j$$ and $$f_u(x+y) = \sum u_n (x+y)^n = \sum_{n} \sum_{i+j=n} \binom{i+j}{i}u_{i+j} x^iy^j$$
The two expressions are equal if and only if the coefficients of $x^iy^j$ are the same for all $i$ and $j$, and that is equivalent to $u_iu_j = \binom{i+j}{i} u_{i+j}$ for all $i$ and $j$.