How can I prove the result in the general case of the binomial series or how can I formally prove in the function I'm given that it is represented by it's Taylor Series?
There are several nice and easy ways. Which is easiest depends on what you already know.
One way uses complex analysis. The function $f_\alpha(z) = (1+z)^\alpha = \exp\bigl(\alpha \log (1+z)\bigr)$ is holomorphic on the open unit disk $\mathbb{D} = \{z\in \mathbb{C} : \lvert z\rvert < 1\}$, hence its Taylor series about $0$ converges locally uniformly to $f_\alpha$ on $\mathbb{D}$.
Now one can show
$$\Biggl(\frac{d}{dz}\Biggr)^n f_\alpha(z) = n!\cdot \binom{\alpha}{n}\cdot (1+z)^{\alpha-n}\tag{1}$$
by induction, and sees that the coefficients of the Taylor series about $0$ are $\binom{\alpha}{n}$.
Another way uses the theory of differential equations. If we let $f_\alpha(x) = (1+x)^\alpha$ and
$$g_\alpha(x) = \sum_{n=0}^\infty \binom{\alpha}{n}x^n,$$
then it is easily seen that $f_\alpha(0) = g_\alpha(0) = 1$. And if we differentiate and then multiply with $(1+x)$, we have
$$(1+x)\cdot\frac{d}{dx}f_\alpha(x) = (1+x)\cdot\frac{d}{dx}(1+x)^\alpha = (1+x)\cdot \alpha (1+x)^{\alpha-1} = \alpha(1+x)^\alpha,$$
and
\begin{align}
(1+x)\cdot \frac{d}{dx} g_\alpha(x)
&= (1+x)\sum_{n=1}^\infty n\binom{\alpha}{n}x^{n-1}\\
&= (1+x) \sum_{n=1}^\infty \alpha\binom{\alpha-1}{n-1}x^{n-1}\\
&= \alpha\cdot(1+x)\sum_{m=0}^\infty \binom{\alpha-1}{m}x^m\\
&= \alpha\Biggl(\sum_{m=0}^\infty \binom{\alpha-1}{m}x^m + \sum_{m=0}^\infty \binom{\alpha-1}{m}x^{m+1}\Biggr)\\
&= \alpha \Biggl(1 + \sum_{m=1}^\infty \biggl(\binom{\alpha-1}{m} + \binom{\alpha-1}{m-1}\biggr)x^m\Biggr)\\
&= \alpha \Biggl(1+\sum_{m=1}^\infty \binom{\alpha}{m} x^m\Biggr)\\
&= \alpha g_\alpha(x).
\end{align}
So the two functions satisfy the same differential equation
$$(1+x)\cdot y' = \alpha y\tag{$\ast$}$$
with the initial condition $y(0) = 1$. By the uniqueness of solutions of $(\ast)$(1), we have $f_\alpha \equiv g_\alpha$ on $\{ x : \lvert x\rvert < 1\}$.
(1) The map $F\colon (x,y) \mapsto \alpha\frac{y}{1+x}$ locally satisfies a Lipschitz condition in $y$ on $(\mathbb{R}\setminus \{-1\})\times \mathbb{R}$, so the Picard-Lindelöf theorem asserts the existence and uniqueness of solutions of $(\ast)$ to any given initial condition $y(x_0) = y_0$ in some neighbourhood of $x_0 \neq -1$.
Not quite as nice and short, but not too bad either is showing that the series converges to $f_\alpha$ by showing that the remainder tends to $0$, where we use the integral form of the remainder term. We assume that $\alpha$ is not a non-negative integer, for in that case, the series is actually a finite sum, and the equality is just the binomial formula.
Like in the complex analysis method, we see by induction that the relation $(1)$ holds, where now we restrict $z$ to be real and $> -1$. Thus the series
$$\sum_{n=0}^\infty \binom{\alpha}{n} x^n$$
is the Taylor series of $f_\alpha$ with centre $0$. The remainder term in the integral form is
$$R_n(x) = \frac{1}{n!} \int_0^x (x-t)^n f_\alpha^{(n+1)}(t)\,dt,$$
using $(1)$, that becomes
$$R_n(x) = (n+1)\binom{\alpha}{n+1}\int_0^x (x-t)^n(1+t)^{\alpha-1-n}\,dt.$$
If $0 \leqslant x < 1$, we let $K = \max \{ 1, (1+x)^\alpha\}$, then $0 < (1+t)^{\alpha-1-n}\leqslant \frac{K}{(1+t)^{n+1}}\leqslant K$ for all $n\in\mathbb{N}$ and $t\in [0,x]$, and we have the estimate
$$\lvert R_n(x)\rvert \leqslant K\cdot(n+1)\left\lvert\binom{\alpha}{n+1}\right\rvert \int_0^x (x-t)^n\,dt = K\left\lvert \binom{\alpha}{n+1}\right\rvert\cdot x^{n+1},$$
and since the series converges for $\lvert y\rvert < 1$, in particular for $y = x$, the terms $\binom{\alpha}{n+1} x^{n+1}$ converge to $0$, so $\lvert R_n(x)\rvert \to 0$.
For $-1 < x < 0$, we write
\begin{align}
\lvert R_n(x)\rvert &= (n+1)\left\lvert \binom{\alpha}{n+1} \int_0^x (x-t)^n(1+t)^{\alpha-1-n}\,dt \right\rvert\\
&= (n+1)\left\lvert \binom{\alpha}{n+1} \int_0^{\lvert x\rvert} (x+t)^n(1-t)^{\alpha-1-n}\,dt\right\rvert\\
&= (n+1) \left\lvert\binom{\alpha}{n+1}\right\rvert \int_0^{\lvert x\rvert} (\lvert x\rvert-t)^n(1-t)^{\alpha-1-n}\,dt\\
&= (n+1) \left\lvert\binom{\alpha}{n+1}\right\rvert \int_0^{\lvert x\rvert} \biggl(\frac{\lvert x\rvert-t}{1-t}\biggr)^n (1-t)^{\alpha-1}\,dt.
\end{align}
Now we note that $t \mapsto \frac{\lvert x\rvert-t}{1-t}$ is monotonically decreasing on $[0,\lvert x\rvert]$, so we obtain the estimate
$$\lvert R_n(x)\rvert \leqslant (n+1)\left\lvert\binom{\alpha}{n+1}\right\rvert\cdot \lvert x\rvert^n \underbrace{\int_0^{\lvert x\rvert} (1-t)^{\alpha-1}\,dt}_C = C\lvert \alpha\rvert\cdot\left\lvert \binom{\alpha-1}{n} x^n\right\rvert.$$
Since also the series $\sum\limits_{n=0}^\infty \binom{\alpha-1}{n}x^n$ converges on the interval $(-1,1)$, it follows that $R_n(x) \to 0$ also for $x\in (-1,0)$.
So we have shown that $R_n(x) \to 0$ for all $x\in (-1,1)$, hence the identity
$$(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n$$
for $x\in (-1,1)$ is proved.
