Suppose that $a_i$ and $b_i$ are the coefficients of the Taylor Series at $\alpha$ of the functions $f$ and $g$. Let $$a_i = \frac{f^{(i)}(\alpha)}{i!}$$ and $$ b_i = \frac{g^{(i)}(\alpha)}{i!}$$
Here is what I did, but I don't know if it's ok. I found that the Taylor Series of $f\cdot g$ of grade $n$ at $\alpha$ has the form $$ R_{n,\alpha}(x) = a_0b_0 + a_0b_1(x-\alpha) + a_0b_2(x-\alpha)^2 + ... + a_0b_n(x-\alpha)^n + a_1b_0(x-a) + ... + a_1b_n(x-\alpha)^{n+1} + ... + a_nb_0(x-\alpha)^{n} + ... + a_nb_n(x-\alpha)^{n+n}$$
So, then I saw that $c_i = a_sb_r = \frac{f^{(s)}(\alpha)}{s!}\cdot\frac{g^{(r)}(\alpha)}{r!}$ with $0 \leq s \leq n$ and $0 \leq r \leq n$ where you can choose any pair of $s$ and $r$. That actually makes $i$ depend on $s$ and $r$, but I couldn't tell who's, for example, $c_0$.
Is there something I'm missing or do I have an error? Help, please. Thank you!
hint
Let $$h=f.g$$ then, by Leibnitz formula, the $ i^{\text{th}}$ coefficient of Taylor series of $ h $ is
$$c_i=\frac{h^{(i)}(\alpha)}{i!}$$
with
$$h^{(i)}(x)=\sum_{k=0}^i\binom{k}{i}f^{(k)}(x)g^{(i-k)}(x)$$ and $$\binom{k}{i}=\frac{i!}{k!(i-k)!}.$$ thus
$$c_i=\sum_{k=0}^i a_kb_{i-k}$$ For example, $$c_2=a_0b_2+a_1b_1+a_2b_0$$