[proofplan]
We prove the iterated derivative formula for each monomial function $x \mapsto x^i$ directly by induction. This formula shows at once that every monomial is smooth and that its derivatives vanish after the exponent. Since $p$ is a finite real linear combination of these monomials, repeated linearity of differentiation gives the corresponding formula for $p$, and when $m>d$ every summand has already vanished.
[/proofplan]
[step:Compute all iterated derivatives of a monomial]
Fix $i \in \{0,\dots,d\}$ and define the monomial function
\begin{align*}
g_i:\mathbb{R}\to\mathbb{R}, \qquad g_i(x)=x^i.
\end{align*}
For every $r \in \mathbb{N} \cup \{0\}$, the derivative $g_i^{(r)}$ exists on $\mathbb{R}$. Moreover, if $0 \le r \le i$, then
\begin{align*}
g_i^{(r)}(x)=\frac{i!}{(i-r)!}x^{i-r}
\end{align*}
for every $x \in \mathbb{R}$, while if $r>i$, then
\begin{align*}
g_i^{(r)}(x)=0
\end{align*}
for every $x \in \mathbb{R}$.
For $r=0$, this is the defining formula $g_i(x)=x^i$. Suppose $0 \le r < i$ and the formula holds for $r$. Differentiating the function $x \mapsto \frac{i!}{(i-r)!}x^{i-r}$ gives
\begin{align*}
g_i^{(r+1)}(x)=\frac{i!}{(i-r)!}(i-r)x^{i-r-1}=\frac{i!}{(i-r-1)!}x^{i-r-1}.
\end{align*}
This proves the formula for all $0 \le r \le i$ by induction. At $r=i$, the formula gives $g_i^{(i)}(x)=i!$, a constant function on $\mathbb{R}$. Therefore its next derivative is zero, and every later derivative is also zero.
[guided]
The goal is to isolate the only computation needed in the proof: what happens when we differentiate a single power $x^i$ repeatedly. Fix $i \in \{0,\dots,d\}$ and define
\begin{align*}
g_i:\mathbb{R}\to\mathbb{R}, \qquad g_i(x)=x^i.
\end{align*}
We claim that for each $r \in \mathbb{N} \cup \{0\}$ the derivative $g_i^{(r)}$ exists on $\mathbb{R}$, and that the formula is
\begin{align*}
g_i^{(r)}(x)=\frac{i!}{(i-r)!}x^{i-r}
\end{align*}
when $0 \le r \le i$, while
\begin{align*}
g_i^{(r)}(x)=0
\end{align*}
when $r>i$.
We prove this by induction through the nonzero derivatives. The case $r=0$ is just the definition of $g_i$, because
\begin{align*}
g_i^{(0)}(x)=g_i(x)=x^i=\frac{i!}{i!}x^i.
\end{align*}
Now suppose $0 \le r < i$ and that the displayed formula for $g_i^{(r)}$ has already been proved. The function $g_i^{(r)}$ is a constant multiple of the power function $x \mapsto x^{i-r}$. Differentiating it gives
\begin{align*}
g_i^{(r+1)}(x)=\frac{i!}{(i-r)!}(i-r)x^{i-r-1}.
\end{align*}
Since
\begin{align*}
\frac{i!}{(i-r)!}(i-r)=\frac{i!}{(i-r-1)!},
\end{align*}
we obtain
\begin{align*}
g_i^{(r+1)}(x)=\frac{i!}{(i-r-1)!}x^{i-r-1}.
\end{align*}
This is exactly the desired formula with $r+1$ in place of $r$.
After $i$ differentiations, the formula becomes
\begin{align*}
g_i^{(i)}(x)=i!
\end{align*}
for every $x \in \mathbb{R}$. This is a constant function, so its derivative is zero:
\begin{align*}
g_i^{(i+1)}(x)=0.
\end{align*}
Once a derivative is the zero function, every later derivative remains the zero function. Hence $g_i^{(r)}(x)=0$ for every $r>i$ and every $x \in \mathbb{R}$.
[/guided]
[/step]
[step:Conclude that each monomial is smooth]
For the fixed index $i \in \{0,\dots,d\}$, the preceding step constructs $g_i^{(r)}$ on all of $\mathbb{R}$ for every $r \in \mathbb{N} \cup \{0\}$. Each such derivative is either a polynomial function or the zero function, hence is continuous on $\mathbb{R}$. Therefore
\begin{align*}
g_i \in C^\infty(\mathbb{R}).
\end{align*}
[/step]
[step:Differentiate the finite polynomial sum term by term]
Since $p$ is the finite real linear combination
\begin{align*}
p=\sum_{i=0}^{d} a_i g_i,
\end{align*}
and each $g_i$ is smooth on $\mathbb{R}$, repeated linearity of differentiation gives, for every $m \in \mathbb{N} \cup \{0\}$ and every $x \in \mathbb{R}$,
\begin{align*}
p^{(m)}(x)=\sum_{i=0}^{d} a_i g_i^{(m)}(x).
\end{align*}
In particular, $p$ has derivatives of all orders on $\mathbb{R}$, and each derivative is a finite sum of continuous functions. Thus
\begin{align*}
p \in C^\infty(\mathbb{R}).
\end{align*}
[/step]
[step:Use the inequality $m>d$ to make every summand vanish]
Let $m \in \mathbb{N}$ satisfy $m>d$. For every index $i \in \{0,\dots,d\}$, we have $m>i$. By the monomial computation, this implies
\begin{align*}
g_i^{(m)}(x)=0
\end{align*}
for every $x \in \mathbb{R}$. Substituting this into the termwise derivative formula gives
\begin{align*}
p^{(m)}(x)=\sum_{i=0}^{d} a_i g_i^{(m)}(x)=\sum_{i=0}^{d} a_i 0=0.
\end{align*}
This holds for every $x \in \mathbb{R}$, completing the proof.
[/step]
