# Use Cauchy's Multiplication Theorem and the Binomial Theorem to prove $\exp(x+y)=\exp(x)\exp(y)$

I am to use Cauchy's Multiplication Theorem and the Binomial Theorem in order to prove

$\exp(x+y)=\exp(x)\exp(y)$

but I have no idea where to begin. All I can think of doing is setting $\exp(x)$ as the sum to infinity of $(x^n)/n!$ and similarly for $\exp(y)$, $(y^n)/n!$

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Well if $x,y\in \mathbb{R}$ then by definition $$\exp(x)\exp(y)=(\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}\frac{y^k}{k!})$$ The Cauchy's Multiplication Theorem tells as that $$\sum_{k=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}=\sum_{k=0}^{\infty}a_k\sum_{k=0}^{\infty}b_k$$ when at least one of the two series of the RHS converge absolutely. In our case we have that $$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}$$ Now because $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, $$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}$$=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k} A straightforward application of the binomial theorem yields the resut: $$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{1}{n!}(x+y)^n=\exp(x+y)$$