I will use the following definitions: (I apologize in advance, my Latex skills are subpar)
$$\binom{\lambda}{n} = \frac{(\lambda).(\lambda -1)...(\lambda-n+1)}{n!}$$ and $$\binom{\lambda}{0} = 1$$ With $\lambda \in \mathbb{C}$
$$(1+X)^{(\lambda)} := \sum_{n=0}^{\infty} \binom{\lambda}{n}\cdot X^n $$
The exercise is as follows;
Prove the following,
$$(1+X)^{\lambda}(1+X)^{\gamma}=(1+X)^{\lambda + \gamma}$$
I attempted to use induction on the nth term of the power series resulting of the product, however I was unable to prove it. Is there a more algebraic aprroach or identity that could crack this problem?
There is a purely algebraic way to show this with formal series, the conditions i needed to prove it were these :
Supposing $\mathbb A$ is a commutative ring that contains $\mathbb Q$ (by that I mean a subring isomorphic to $\mathbb Q$, because if there is one then it is unique and the isomorphism is unique as well) then the structure is enough to define binomial coefficients : $$\forall \alpha \in \mathbb A, \, \forall n \in \mathbb Z, \, \binom{\alpha}{n} := \left\{ \begin{aligned} &\frac{\prod_{i=0}^{n-1} (\alpha-i)}{n!} & \text{ if } n \geq 0 \\ &0 & \text{ otherwise} \end{aligned} \right.$$ Then we have Vandermonde's identity (the one you're trying to prove) $$\forall \alpha, \beta \in \mathbb A, \; \forall n \in \mathbb N, \; \sum_{i=0}^n \binom{\alpha}{i} \binom{\beta}{n-i} = \binom{\alpha+\beta}{n}$$
Proof :
We'll be using $$\exp := \sum_{n \geq 0} \frac{1}{n!} X^n$$ $$U := \sum_{n \geq 1} \frac{(-1)^{n+1}}{n} X^n \text{ (basically }\ln(1+X)\text{)}$$ $$\forall \alpha \in \mathbb A, \; (1+X)^\alpha := \exp \circ (\alpha U)$$
First, we show that with this new definition of $(1+X)^\alpha$, we have $$(1+X)^0 = 1$$ $$(1+X)^1 = 1+X$$ $$\forall \alpha, \beta \in \mathbb A, \; (1+X)^\alpha (1+X)^\beta = (1+X)^{\alpha+\beta}$$ So our new notation doesn't conflict with integer powers of $1+X$
Second, we show that the definition I've given to $(1+X)^\alpha$ matches with the other definition, that uses binomial coefficients : \begin{align*} \forall \alpha \in \mathbb A, \; ((1+X)^\alpha)' &= \alpha U' \cdot \exp' \circ (\alpha U) \\ &= \alpha (1+X)^{-1} (1+X)^\alpha \\ &= \alpha (1+X)^{\alpha-1} \end{align*} We get the $n$-th derivative by induction : $$\forall \alpha \in \mathbb A, \; \forall n \in \mathbb N, \; ((1+X)^\alpha)^{(n)} = \left(\prod_{i=0}^{n-1} \alpha - i \right) (1+X)^{\alpha-n}$$ and the coefficients of $(1+X)^\alpha$ : \begin{align*} \forall n \in \mathbb N, \; [X^n](1+X)^\alpha &= \frac{1}{n!}[X^0]((1+X)^\alpha)^{(n)} \\ &= \frac{\prod_{i=0}^{n-1} (\alpha - i)}{n!} [X^0] (1+X)^{\alpha-n} \\ &= \binom{\alpha}{n} [X^0] \exp \\ &= \binom{\alpha}{n} \end{align*} and the identity $$\forall \alpha, \beta \in \mathbb A, \; (1+X)^\alpha (1+X)^\beta = (1+X)^{\alpha+\beta}$$ yields $$\forall \alpha, \beta \in \mathbb A, \; \forall n \in \mathbb N, \; \sum_{i=0}^n \binom{\alpha}{i} \binom{\beta}{n-i} = \binom{\alpha+\beta}{n}$$