I am trying to show that the function that satisfies $f^\prime(x)=f(x)$ with $f(0)=1$ behaves in an exponential way (in other words, I want to justify writing it as $e^x$). I need to show that: $$ f(a)f(b) =f(a+b). $$ We have: $$ f(a)f(b) = \left( \sum_{k=0}^{\infty} \dfrac{a^k}{k!} \right)\left( \sum_{k=0}^{\infty} \dfrac{b^k}{k!} \right), $$ and using a Cauchy product, $ f(a)f(b) = \sum_{n=0}^{\infty} c_n $ where:
$$ c_n = \sum_{k=0}^{n} \dfrac{a^k}{k!} \dfrac{b^{n-k}}{(n-k)!} = \dfrac{b^n}{n!} \sum_{k=0}^{n} \left[\dfrac{a}{b}\right]^k \binom{n}{k}. $$
I am stuck here, because I do not know how to evaluate the sum with the binomial coefficient that includes a power of $k$. Any help would be appreciated.
By the Binomial Theorem, the sum from $k=0$ to $n$ is $(1+\alpha)^n$.