$1,(x-5),(x-5)^2,...,(x-5)^m$ as a basis for $\mathcal{P}_m(\mathbf{R})$ Is the following Proof Correct?

Theorem. Given that $m$ is a positive integer. the following list of vectors is a basis for $\mathcal{P}_m(\mathbf{R})$.
  $$1,(x-5),(x-5)^2,\dots ,(x-5)^m$$

Proof. The list presented above has a length of $m+1 = \dim\mathcal{P}_m(\mathbf{R})$ consequently it is sufficient for us to demonstrate that the above list is linearly independent.
Assume that $c_0+c_1(x-5)+c_2(x-5)^2+\cdots+c_m(x-5)^m = 0$, where $c_1,c_2,\dots,c_m\in\mathbf{R}$ and for convenience let $p\in\mathcal{P}_m(\mathbf{R})$ represent the L.H.S of the above equation then $p^{(m)}(x) = c_m\cdot m! = 0$ which implies that $c_m = 0$ consequently $p(x) = c_0+c_1(x-5)+c_2(x-5)^2+\cdots+c_{m-1}(x-5)^{m-1}$ repeating the same process we see that $p^{m-1}(x) = c_{m-1}(m-1)! = 0$ which implies that $c_{m-1} = 0$ continuing in this manner we can show that $c_{m} = c_{m-1} = \cdots = c_1 = 0$ and evaluating $p$ at $x=5$ implies that $c_0=0$.
$\blacksquare$
 A: Your proof is correct. 
I would approach the question differently. 
It suffices to show that this set of
 $$1,(x-5),(x-5)^2,\dots ,(x-5)^m$$ span $$ 1, x, x^2,..., x^m$$  
Note that $$ 1=1$$, $$x=5(1)+ (x-5) $$,
$$ x^2 = 25(1)+10(x-5)+ (x-5)^2 $$ 
Which could be achieved by Taylor polynomial of $x^k$ around $a=5.$
A: Yes it is correct and you can conclude directly from here since
$$p=c_0+c_1(x-5)+c_2(x-5)^2+\cdots+c_m(x-5)^m = 0$$
is true if and only if $p$ is the zero polynomial with $c_i=0$ $\forall i$.
As an alternative we could shift from $x-5$ to y and observe that we obtain the standard basis $1,y,..,y^m$.
A: If you know that $\{\color{blue}{1,x,x^2,\ldots,x^m}\}$ forms a basis for for $\mathcal{P}_m(\mathbf{R})$, then it's enough to note that the following gives the explicit change of basis to $\{\color{red}{1,(x-5),(x-5)^2,\ldots,(x-5)^m}\}\,$:
$$
\color{blue}{x^k} = \big((x-5)+5\big)^k = \sum_{j=0}^k 5^{k-j}\binom{k}{j} \cdot \color{red}{(x-5)^j} \quad\quad\quad k = 0,1,2,\ldots,m
$$
