Let $f:\mathbb R\to\mathbb R$ be an $n+1$-times differentiable function. Taylor's Theorem states that for each $x\in\mathbb R$,
$$
f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(n)}(0)}{n!}x^n+\frac{f^{(n+1)}(\lambda)}{(n+1)!}x^{n+1} \tag{*}\label{*}
$$
where $\lambda$ is strictly in between $0$ and $x$.
Note that in the case $n=0$, this is simply a restatement of the mean value theorem. However, the proof for general $n$ found in many textbooks is a rather messy inductive proof.
My question is: is there an intuitive reason why $\eqref{*}$ holds? The mean value theorem has an appealing geometric interpretation, that at some point the instantaneous rate of change over an interval has to equal the average rate of change. I wonder if there is an analogous interpretation of $\eqref{*}$.
To clarify, it has already been discussed on MSE why we ought to have
$$
f(x)\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{(n)}(0)}{n!}x^n \, ,
$$
for $x$ close to $0$, and I do find this rather intuitive. The reason I find Taylor's theorem somewhat unintuitive is because it is unclear why the remainder term can be expressed in the above form.
Best Answer
You wrote "the proof for general $n$ found in many textbooks is a rather messy inductive proof." I agree, but it turns out there is a much simpler argument relying on nothing more than Rolle's theorem, so it's depends on nothing more than a special case of the Mean Value Theorem.
I discovered this yesterday in a comment by Pieter-Jan De Smet on the blog page of Gowers about the Lagrange form of the remainder here, where the simpler proof comes from a textbook around 40 years ago. Or see Theorem 1.1 and its proof in Section 2 here, which is my account of two different forms of the remainder. I used to have an inductive proof there for the Lagrange form of the remainder, but I rewrote it to present the simpler argument from that blog comment.