For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term

Base case, claim 2 at $n = 1$. By [L1], $x^{1} = x^{0}x = 1 \cdot x = x$, so $p_1$ is the identity. Fix $c \in \mathbb{R}$; for every $x \ne c$ the difference quotient is $(x - c)/(x - c) = 1$, so it is the constant function $1$ on $\mathbb{R} \setminus \{c\}$, and by [L4] and [L8] its limit at $c$ is $1$. Since $\iota(1)c^{0} = 1 \cdot 1 = 1$ by [L1] and [L2], claim 2 holds at $n = 1$.

Inductive hypothesis. Fix a natural $n \ge 1$ and assume that $p_n$ is differentiable at every $c \in \mathbb{R}$ with $p_n'(c) = \iota(n)c^{\,n-1}$.

Claim 1. By [L1] the function $p_0$ is the constant function $1$. Fix $c \in \mathbb{R}$; for every $x \ne c$ its difference quotient is $(1-1)/(x-c) = 0$, the constant function $0$ on $\mathbb{R}\setminus\{c\}$, whose limit at $c$ is $0$ by [L4] and [L8]. So $p_0$ is differentiable at every $c$ with $p_0'(c) = 0$.

Successor step. Let $c \in \mathbb{R}$. By [L1], $p_{n+1}(x) = x^{n+1} = x^{n}x = p_n(x)\,p_1(x)$ for every $x$. Both factors are differentiable at $c$, by step 1.2 and step 1.1, so the product rule of [L3] gives that $p_{n+1}$ is differentiable at $c$ with $p_{n+1}'(c) = p_n'(c)p_1(c) + p_n(c)p_1'(c) = \iota(n)c^{\,n-1}c + c^{n} \cdot 1$. Now $c^{\,n-1}c = c^{\,(n-1)+1} = c^{n}$ by [L1], so the right-hand side is $\iota(n)c^{n} + c^{n} = \bigl(\iota(n)+1\bigr)c^{n} = \iota(n+1)c^{n}$ by [L2].

Claim 2. Steps 1.1 and 2.1 are the base case and the successor step of an induction over the naturals $n \ge 1$, so by [L5] the function $p_n$ is differentiable at every $c \in \mathbb{R}$ with $p_n'(c) = \iota(n)c^{\,n-1}$, for every natural $n \ge 1$.

Claim 3. Let $n \ge 1$. The set $\{\, x \in \mathbb{R} : p_n(x) \ne 0 \,\}$ is exactly $A_0$: a nonzero $x$ has $x^{n} \ne 0$ by [L7], and $0^{n} = 0$ by [L1]. Fix $c \in A_0$. The constant function $1$ on $\mathbb{R}$ is differentiable at $c$ with derivative $0$ by step 1.3, and $p_n$ is differentiable at $c$ with derivative $\iota(n)c^{\,n-1}$ by step 3.1, with $p_n(c) = c^{n} \ne 0$. So the quotient rule of [L3] applies: the function $x \mapsto 1/x^{n}$ on $A_0$, which by [L1] and [L7] is $m_n$, is differentiable at $c$ with derivative $\bigl(0 \cdot c^{n} - 1 \cdot \iota(n)c^{\,n-1}\bigr)/(c^{n})^{2} = -\iota(n)\,c^{\,n-1}\bigl(c^{\,2n}\bigr)^{-1} = -\iota(n)\,c^{\,n-1-2n} = -\iota(n)\,c^{\,-n-1}$, where $(c^{n})^{2} = c^{\,2n}$ and $c^{\,n-1}c^{\,-2n} = c^{\,-n-1}$ are [L7].

Claim 4, by a second induction on $n$. Fix $c \in \mathbb{R}$ and the sequence $a$. At $n = 0$ the sum $P(x) = \sum_{j<0} a_j x^{j}$ is $0$ for every $x$ by [L6], so $P$ is the constant function $0$ and, as in step 1.3, $P'(c) = 0 = \sum_{j<0} e_j$. Suppose the claim holds at $n$, and let $P_{n+1}(x) := \sum_{j<n+1} a_j x^{j}$. By [L6], $P_{n+1}(x) = P_n(x) + a_n x^{n}$ for every $x$, where $P_n(x) = \sum_{j<n} a_j x^{j}$. The function $x \mapsto a_n x^{n}$ is differentiable at $c$ with derivative $e_n$: for $n = 0$ it is the constant $a_n$, of derivative $0 = e_0$, by step 1.3 and the scalar rule of [L3]; for $n \ge 1$ it is the scalar multiple $a_n p_n$, of derivative $a_n \iota(n)c^{\,n-1} = e_n$, by step 3.1 and the scalar rule of [L3]. The inductive hypothesis gives $P_n'(c) = \sum_{j<n} e_j$, so the sum rule of [L3] gives that $P_{n+1}$ is differentiable at $c$ with $P_{n+1}'(c) = \sum_{j<n} e_j + e_n = \sum_{j<n+1} e_j$ by [L6]. By [L5] claim 4 holds for every $n \in \mathbb{N}$.

All four claims are established: claim 1 by step 1.3, claim 2 by step 3.1, claim 3 by step 4.1 and claim 4 by step 4.2.