Most Calculus books do not rigorously define and prove the main properties of basic functions such as $\sqrt{x}$, $\sin(x)$, and $e^x$. The reason is that this is not simple. For instance, one may ask how a computer or calculator actually evaluates these functions. One way is by using power series, which usually appear later in Calculus. There are, however, many other approaches, some of them even more elementary.
I will give a deduction of the Binomial formula for $(1+x)^{1/2}$ using only basic facts about power series (we will not use Taylor series or the logarithm explicitly). The idea is to invert the argument: we first define a function $f_{\alpha}$ through the generalized binomial coefficients for $\alpha\in \mathbb{R}$, and then we prove at the end that $f_{1/n}(x) = (1+x)^{1/n}$ for $|x|<1$ and $n \in \mathbb{N}^\star$.
Define
$$
\binom{\alpha}{k} = \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(k-1))}{k!},
\qquad k \in \mathbb{N}, \ \alpha \in \mathbb{R}.
$$
Note that if $\alpha \in \mathbb{N}$ and $\alpha \geq k$ this coincides with the usual definition of the binomial coefficients.
Define
$$
f_\alpha(z) = \sum_{k=0}^\infty \binom{\alpha}{k} z^k.
$$
Our goal is to show that
(1) This function is well-defined for $|z|<1$ (the power series converges) and $f_\alpha(z) > 0$ for $|z|<1$. In particular derivatives of arbitrary order exist on this interval.
(2) If $n \in \mathbb{N}^\star$ then $f_n(z) = (1+z)^n$.
(3) If $n \in \mathbb{N}^\star$ then $(f_{1/n}(z))^n = 1+z$.
This allows us to define for $x \in (0,2)$
$$
x^{1/2} := f_{1/2}(x-1)
$$
and to prove that $x^{1/2}$ is differentiable of arbitrary order, and to use the chain rule to calculate all its derivatives at $x=1$ without circular arguments.
Step 1. Radius of convergence.
If $\alpha \notin \mathbb{N}$ apply the ratio test to conclude that the radius of convergence is $R=1$.
If $\alpha \in \mathbb{N}$ then the product $\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(k-1))$ vanishes once $k>\alpha$. Therefore $f_\alpha$ is a polynomial. In this case the radius of convergence is $+\infty$.
Step 2. Functional relation.
$$
(1+z) f_\alpha(z) = f_{\alpha+1}(z), \qquad |z|<1.
$$
Hint: expand both sides as power series and check that the coefficients match.
Step 3. For $n \in \mathbb{N}^\star$ we have
$$
f_n(z) = (1+z)^n.
$$
Hint: use induction on $n$, Step 2, and the fact that $f_1(z) = 1+z$. This gives the Binomial Theorem for $(1+z)^n$ with $n \in \mathbb{N}$.
Step 4. Derivative relation.
$$
f_\alpha'(z) = \alpha f_{\alpha-1}(z), \qquad |z|<1, \ \alpha \neq 0.
$$
Hint: if $\alpha \notin \mathbb{N}$ differentiate the series term by term and compare coefficients. If $\alpha \in \mathbb{N}$ use Step 3.
Step 5. Differential equation.
$$
(1+z) f_\alpha'(z) = \alpha f_\alpha(z), \qquad |z|<1.
$$
This is a first-order differential equation satisfied by $f_\alpha$.
Hint: combine Steps 2 and 4.
Step 6. If $\alpha \in \mathbb{N}^\star$ then
$$
(f_{1/\alpha}(z))^\alpha = 1+z, \qquad |z|<1.
$$
In particular, taking $\alpha=2$ we obtain $f_{1/2}(z) = \sqrt{1+z}$.
Proof: define
$$
h(z) = (f_{1/\alpha}(z))^\alpha.
$$
By Step 5 we get
$$
h'(z) = \alpha (f_{1/\alpha}(z))^{\alpha-1} \cdot \frac{1}{\alpha} \frac{\left(f_{1/\alpha}(z)\right)^\alpha}{1+z} = \frac{h(z)}{1+z}.
$$
Thus
$$
(1+z) h'(z) - h(z)=0,
$$
so since $h(0)=1$ and
$$\left(\frac{h(z)}{1+z} \right)'= \frac{(1+z) h'(z) - h(z)}{(1+z)^2} =0$$
we conclude that $h(z)=1+z$.
Step 7. For all $\alpha\in \mathbb{R}^\star$ we have $$f_\alpha(z)f_{-\alpha}(z)=1, \qquad |z|<1.$$
Hint: Differentiate the left-hand side using the product rule and Step 5, and conclude that its derivative is zero.
In particular $f_\alpha(z)\neq 0$ for $|z|< 1$ and since $f_\alpha$ is continuous on $(-1,1)$ and $f_\alpha(0)=1> 0$ it follows that $f_\alpha(z) > 0$ for $|z|< 1$.
