Prove that $(1 + z)^x \cdot (1+z)^y = (1+z)^{(x+y)}$ for formal power series Imagine that I define a function by formal power series:
$$ (1 + z)^x := \sum_{n = 0}^{\infty} {x \choose n} z^n $$
where ${x \choose n} := \frac{x \cdot (x - 1) \ldots \cdot (x - n + 1)}{n!}$
How can I prove that $(1 + z)^x \cdot (1+z)^y = (1+z)^{(x+y)}$?
 A: It is clear that that equality holds if $x,y\in\mathbb Z_+$. On the other hand,\begin{align}(1+z)^x(1+z)^y&=\left(\sum_{n=0}^\infty\binom xnz^n\right)\left(\sum_{n=0}^\infty\binom ynz^n\right)\\&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom xk\binom y{n-k}\right)z^n\end{align}and therefore your equality is equivalent to the assertion than, for each $n\in\mathbb Z_+$,$$(\forall x,y\in\mathbb C):\binom{x+y}n=\sum_{k=0}^n\binom xk\binom y{n-k}.$$So, define$$P_n(x,y)=\binom{x+y}n-\sum_{k=0}^n\binom xk\binom y{n-k}.$$You want to prove that each $P_n$ is the null polynomial. This is the same thing as asserting that, for each $x\in\mathbb C$, $y\mapsto P_n(x,y)$ is the null polynomial. Since this is a polynomial function in one variable, it is enough to prove that $(\forall k\in\mathbb Z_+):P_n(x,k)=0$. Fix $k\in\mathbb Z_+$. You wish to prove that $x\mapsto P_n(x,k)$ is the null map. Again, since it is a polynomial map, it will be enough to pro that $(\forall l\in\mathbb Z_+):P_n(l,k)=0$. But I noted before that the statement that you want to prove holds for non-negative integers.
A: Another approach is to use formal differentiation. Over the ring of formal power series in $z$, $f(z)=(1+z)^x$ is the unique solution of the
differential equation $zf'(z)=xf(z)$ with initial condition $f(0)=1$. The differential equation is just a recurrence on the coefficients of the formal power
series, and the proof the $n$-th coefficient is $\binom xn$ follows by induction.
Then $F(x)=(1+z)^x(1+z)^y$ has $F(0)=1$ and has
$$zF'(z)=x(1+z)^x(1+z)^y+(1+z)^xy(1+z)^y=(x+y)F(z)$$
so that $F(z)$ must equal $(1+z)^{x+y}$.
