Here is Prob. 18, Chap. 5, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $f$ is a real function on $[a, b]$, $n$ is a positive integer, and $f^{(n-1)}$ exists for every $t \in [a, b]$. Let $\alpha$, $\beta$, and $P$ be as in Taylor's theorem (5.15). Define $$ Q(t) = \frac{ f(t) - f(\beta) }{ t- \beta} $$ for $t \in [a, b]$, $t \neq \beta$, differentiate $$ f(t) - f(\beta) = (t-\beta) Q(t) $$ $n-1$ times at $t = \alpha$, and derive the following version of Taylor's theorem: $$ f(\beta) = P(\beta) + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta - \alpha)^n. $$
And, here is Theorem 5.15 in Baby Rudin, 3rd edition:
Suppose $f$ is a real function on $[a, b]$, $n$ is a positive integer, $f^{(n-1)}$ is continuous on $[a, b]$, and $f^{(n)}(t)$ exists for every $t \in (a, b)$. Let $\alpha$, $\beta$ be distinct points of $[a, b]$, and define $$ P(t) = \sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!} \left( t-\alpha \right)^k.$$ Then there exists a point $x$ between $\alpha$ and $\beta$ such that $$ f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!} (\beta - \alpha )^n.$$
An Attempt:
For all $t \in [a, b]$, we have $$ \begin{align} f(t) - f(\beta) &= ( t-\beta) Q(t), \tag{1} \\ f^\prime(t) &= Q(t) + (t-\beta) Q^\prime(t), \tag{2} \\ f^{\prime\prime}(t) &= 2Q^\prime(t) + (t-\beta) Q^{\prime\prime}(t), \tag{3} \\ f^{(3)}(t) &= 3 Q^{\prime\prime}(t) + (t-\beta)Q^{(3)}(t), \tag{4} \\ f^{(4)}(t) &= 4 Q^{(3)}(t) + (t-\beta) Q^{(4)}(t), \tag{5} \\ \cdots &= \cdots \\ f^{(n-1)}(t) &= (n-1) Q^{(n-2)}(t) + (t-\beta) Q^{(n-1)}(t). \tag{*} \end{align} $$ So, for $t = \alpha$, the above chain of equations yields $$ \begin{align} & \qquad f(\beta) \\ &= f(\alpha) + Q(\alpha) (\beta - \alpha ) \qquad \mbox{ [ using (1) ] } \\ &= f(\alpha) + \left[ f^\prime(\alpha) + (\beta - \alpha) Q^\prime(\alpha) \right] (\beta - \alpha ) \\ & \qquad \qquad \mbox{ [ using (2) ] } \\ &= f(\alpha) + f^\prime(\alpha) (\beta - \alpha) + Q^\prime(\alpha) (\beta - \alpha)^2 \\ &= f(\alpha) + f^\prime(\alpha) (\beta - \alpha) \\ & \qquad + \left[ \frac{1}{2} \left( f^{\prime\prime}(\alpha) + (\beta - \alpha) Q^{\prime\prime}(\alpha) \right) \right] (\beta - \alpha)^2 \\ & \qquad \qquad \mbox{ [ using (3) ] } \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{Q^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^3 \\ &= f(\alpha) + f^\prime(\alpha) (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2} (\beta - \alpha)^2 + \frac{\frac{1}{3} \left( f^{(3)}(\alpha) + (\beta - \alpha) Q^{(3)}(\alpha) \right) }{2} (\beta - \alpha)^3 \qquad \mbox{ [ using (4) ] } \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{Q^{(3)}(\alpha)}{3!} (\beta-\alpha)^4 \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{ \frac{1}{4} \left( f^{(4)}(\alpha) + (\beta - \alpha) Q^{(4)}(\alpha) \right) }{3!} (\beta-\alpha)^4 \\ & \qquad \qquad \mbox{ [ using (5) ] } \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{f^{(4)}(\alpha)}{4!} (\beta - \alpha)^4 + \frac{Q^{(4)}(\alpha)}{4!} (\beta-\alpha)^5 \\ &= \cdots \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 \\ & \qquad + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{f^{(4)}(\alpha)}{4!} (\beta - \alpha)^4 + \cdots + \frac{f^{(n-1)}(\alpha)}{(n-1)!} (\beta-\alpha)^{n-1} + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta - \alpha)^n \\ &= P(\beta) + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta - \alpha)^n, \end{align} $$ as required.
Is this proof correct? If so, then is it rigorous enough for Rudin as well?
