I've found this proof which I am quite proud of. Am I missing anything?
Theorem: $$\sum_{k=0}^{n}{n\choose k}=2^n$$ Proof:
Let $\beta\in\{t\in\Bbb R:t>0\}$, and $f(x)=x^\beta$ be continuous on some interval $I\subseteq\Bbb R$, where $1\in I$. Assume that $D^mf(x)$ (the $m$-th derivative of $f(x)$), is continuous on $I$ for any $m\in\Bbb N_0$.
From the definition of Taylor series, $$f(x)=\sum_{k\geq0}\frac{D^kf(1)}{k!}(x-1)^k$$ Note: for the sake of clarity $\sum_{k\geq0}=\sum_{k=0}^{\infty}$
From the power rule, $$D^kf(x)=(\beta)_kx^{\beta-k}$$ $$\therefore D^kf(1)=(\beta)_k$$ Where $(\beta)_k=\prod_{i=1}^k(\beta-i+1)$ is the falling factorial
Therefore: $$f(x)=x^\beta=\sum_{k\geq0}\frac{(\beta)_k}{k!}(x-1)^k$$ $$\therefore x^\beta=\sum_{k\geq0}{\beta\choose k}(x-1)^k$$ $$\therefore 2^\beta=\sum_{k\geq0}{\beta\choose k}(2-1)^k$$ $$\therefore 2^\beta=\sum_{k\geq0}{\beta\choose k}$$ Let $n\in\Bbb N_0$. Evaluating the series at $\beta=n$ gives $$2^n=\sum_{k\geq0}{n\choose k}$$ Recall the identity that, for all $k\in\Bbb Z$, where $k>n$, ${n\choose k}=0$, because $(n)_k=0$ for such values of $k$.
Therefore $$2^n=\sum_{k=0}^{n}{n\choose k}$$ QED.