I am stuck at this in proof of Taylor's theorem in Rudin PMA:
How to get $g^{(n)}(t)=f^{(n)}(t)-n!M$, if $g(t)=f(t)-\sum\limits_{k=0}^{n-1}\dfrac{f^{(k)}(\alpha)}{k!}(t-\alpha)^k-M(t-\alpha)^n$?
I tried a lot, but I don;'t know where I am stuck:
$$g^{(n)}(t)=f^{(n)}(t)-f'(\alpha)-f''(\alpha)-\dots-\dfrac{f^{(n)}(\alpha)}{k!}k(t-\alpha)^{k-1}-nM(t-\alpha)^{n-1}$$
The first term $f^{(n)}(t)$ comes from the $n$-th derivative of $f(t)$.
Observe that $\dfrac{f^{(k)}(\alpha)}{k!}$ is the coefficient of the polynomial $(t-\alpha)^k$ of degree $k < n$, after differentiating $n$-times, the sum $\sum\limits_{k=0}^{n-1}\dfrac{f^{(k)}(\alpha)}{k!}(t-\alpha)^k$ vanishes.
The last term $-n!M$ comes from the $n$-th derivative of $-M(t-\alpha)^n$.