[proofplan]
We first verify that every power $T^k$ is an $F$-linear endomorphism of $V$, using the identity map for $k=0$ and induction on $k$ for the positive powers. Once this is known, each scalar multiple $a_kT^k$ is again an $F$-[linear map](/page/Linear%20Map) from $V$ to $V$. Finally, we prove linearity of the finite sum defining $p(T)$ directly by evaluating it on $\alpha v+\beta w$ and distributing through the sum.
[/proofplan]
[step:Show that every power $T^k$ is an $F$-linear endomorphism]
For $k=0$, the map $T^0=\operatorname{id}_V:V\to V$ is $F$-linear, since for all $\alpha,\beta\in F$ and all $v,w\in V$,
\begin{align*}
\operatorname{id}_V(\alpha v+\beta w)=\alpha v+\beta w=\alpha\operatorname{id}_V(v)+\beta\operatorname{id}_V(w).
\end{align*}
We prove by induction that $T^k\in \operatorname{End}_F(V)$ for every integer $k\geq 0$. The case $k=0$ was just proved. Assume that $T^k:V\to V$ is $F$-linear for some integer $k\geq 0$. By definition, $T^{k+1}=T\circ T^k:V\to V$. For all $\alpha,\beta\in F$ and all $v,w\in V$, the linearity of $T^k$ and then the linearity of $T$ give
\begin{align*}
T^{k+1}(\alpha v+\beta w)=T(T^k(\alpha v+\beta w))=T(\alpha T^k(v)+\beta T^k(w)).
\end{align*}
Applying linearity of $T$ to the last expression gives
\begin{align*}
T^{k+1}(\alpha v+\beta w)=\alpha T(T^k(v))+\beta T(T^k(w))=\alpha T^{k+1}(v)+\beta T^{k+1}(w).
\end{align*}
Thus $T^{k+1}$ is $F$-linear. By induction, $T^k\in \operatorname{End}_F(V)$ for every integer $k\geq 0$.
[guided]
The first point is that the expression $p(T)$ only makes sense as an endomorphism-valued expression if every power $T^k$ is itself a map from $V$ to $V$ and is $F$-linear. We start with $k=0$. By definition, $T^0=\operatorname{id}_V:V\to V$. For arbitrary scalars $\alpha,\beta\in F$ and arbitrary vectors $v,w\in V$, the identity map satisfies
\begin{align*}
\operatorname{id}_V(\alpha v+\beta w)=\alpha v+\beta w.
\end{align*}
Since $\operatorname{id}_V(v)=v$ and $\operatorname{id}_V(w)=w$, this becomes
\begin{align*}
\operatorname{id}_V(\alpha v+\beta w)=\alpha\operatorname{id}_V(v)+\beta\operatorname{id}_V(w).
\end{align*}
Therefore $T^0=\operatorname{id}_V$ is $F$-linear.
Now fix an integer $k\geq 0$ and assume as the induction hypothesis that $T^k:V\to V$ is $F$-linear. The next power is defined by composition:
\begin{align*}
T^{k+1}=T\circ T^k.
\end{align*}
This is a map $V\to V$ because $T^k$ maps $V$ into $V$ and $T$ also maps $V$ into $V$. To prove that $T^{k+1}$ is $F$-linear, take arbitrary $\alpha,\beta\in F$ and arbitrary $v,w\in V$. First use the definition of composition:
\begin{align*}
T^{k+1}(\alpha v+\beta w)=T(T^k(\alpha v+\beta w)).
\end{align*}
The induction hypothesis says that $T^k$ is $F$-linear, so
\begin{align*}
T^k(\alpha v+\beta w)=\alpha T^k(v)+\beta T^k(w).
\end{align*}
Substituting this into the previous expression gives
\begin{align*}
T^{k+1}(\alpha v+\beta w)=T(\alpha T^k(v)+\beta T^k(w)).
\end{align*}
Since $T$ is $F$-linear by the hypothesis $T\in\operatorname{End}_F(V)$, we may distribute $T$ across this linear combination:
\begin{align*}
T(\alpha T^k(v)+\beta T^k(w))=\alpha T(T^k(v))+\beta T(T^k(w)).
\end{align*}
Using again the definition $T^{k+1}=T\circ T^k$, this becomes
\begin{align*}
T^{k+1}(\alpha v+\beta w)=\alpha T^{k+1}(v)+\beta T^{k+1}(w).
\end{align*}
Thus $T^{k+1}$ is $F$-linear. The induction proves that $T^k$ is an $F$-linear endomorphism of $V$ for every integer $k\geq 0$.
[/guided]
[/step]
[step:Check that each scalar multiple $a_kT^k$ is $F$-linear]
For each integer $k$ with $0\leq k\leq m$, define
\begin{align*}
S_k:V&\to V, & v&\mapsto a_kT^k(v).
\end{align*}
Since $T^k$ is $F$-linear by the previous step, for all $\alpha,\beta\in F$ and all $v,w\in V$,
\begin{align*}
S_k(\alpha v+\beta w)=a_kT^k(\alpha v+\beta w).
\end{align*}
Using linearity of $T^k$ and distributivity in the [vector space](/page/Vector%20Space) $V$, we obtain
\begin{align*}
S_k(\alpha v+\beta w)=a_k(\alpha T^k(v)+\beta T^k(w))=\alpha a_kT^k(v)+\beta a_kT^k(w)=\alpha S_k(v)+\beta S_k(w).
\end{align*}
Thus $S_k$ is $F$-linear for each $k$ with $0\leq k\leq m$.
[/step]
[step:Compute the finite sum to prove that $p(T)$ is linear]
Let $\alpha,\beta\in F$ and let $v,w\in V$. By definition of $p(T)$,
\begin{align*}
p(T)(\alpha v+\beta w)=\sum_{k=0}^{m}a_kT^k(\alpha v+\beta w).
\end{align*}
Since each $T^k$ is $F$-linear, this becomes
\begin{align*}
p(T)(\alpha v+\beta w)=\sum_{k=0}^{m}a_k(\alpha T^k(v)+\beta T^k(w)).
\end{align*}
Using distributivity in $V$ and the fact that finite sums in a vector space may be separated term by term, we get
\begin{align*}
p(T)(\alpha v+\beta w)=\alpha\sum_{k=0}^{m}a_kT^k(v)+\beta\sum_{k=0}^{m}a_kT^k(w).
\end{align*}
By the definition of $p(T)$ on $v$ and on $w$, the right-hand side is
\begin{align*}
\alpha p(T)(v)+\beta p(T)(w).
\end{align*}
Therefore
\begin{align*}
p(T)(\alpha v+\beta w)=\alpha p(T)(v)+\beta p(T)(w).
\end{align*}
Because $\alpha,\beta\in F$ and $v,w\in V$ were arbitrary, $p(T):V\to V$ is $F$-linear. Hence $p(T)\in \operatorname{End}_F(V)$.
[/step]
