question about solution of $1,(x-5),(x-5)^2,...,(x-5)^m$ as a basis for $\mathcal{P}_m(\mathbf{R})$ axler 3.F.8  here is another answer for but I have problem for understanding highlighted part:

first why $\varphi_j(0)$??
second isn't any problem for $j=0$?
 A: The $j$th derivative of $$a_0+a_1(x-5)+\ldots +a_m(x-5)^m$$ is $$j!\cdot a_j+b_{j+1}(x-5)+b_m(x-5)^{(m-j)}$$ (for some constants $b_{j+1},\ldots ,b_m$ that we don't need to know, but that one can easily calculate), so when you evaluate it a $x=5$, you get $j!\cdot a_j$, and then $\varphi_j(p)=a_j$. On the other hand, exactly as in the question you have cited, here we are supposing that $p(x)=a_0+a_1(x-5)+\ldots +a_m(x-5)^m$ is the null vector, so $\varphi_j(p)=\varphi_j(0)=0$.
When $j=0$, conventionally one have to take $p^{(0)}$ as $p$, see for example here (find "zeroth derivative" in the first paragraph) or here (paragraph 1.2, "Taylor series").
Note that actually there is a typo, the coefficients $a_0, \ldots ,a_m$ are in $\mathbb{R}$ not in $\mathbb{F}$, but actually you can consider also polynomials over any infinite field (viewed as an extension of $\mathbb{Q}$).
A: We have an equality of polynomials
$$
a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m = 0
$$
Applying $\varphi_j$ to both sides, we keep the equality and get
$$
\varphi_j(a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m) = \varphi_j(0)
$$
The main difficulty here is actually recognizing that the left-hand side is equal to $a_j$, not that the left-hand side and right-hand side are equal.
As for what happens when $j = 0$, there is no issue. Differentiating $0$ times is the same as just not doing anything, so $p^{(0)}$ is just $p$. And $0!$ is $1$. So $\varphi_0(p)$ turns out to just be $p(5)$. Which for our particular (left-hand side) polynomial turns out to be just $a_0$.
