Error in Taylor polynomial

Let $f$ be a function, $b<d$, $n\in\mathbb{N}$, $f^{(n)}$ is continuous on $[b,d]$, $f^{(n+1)}$ exists on $(b,d)$. The remainder of its $n$-th degree Taylor polynomial at $a\in [b,d]$ is $$R_n(x):=f(x)-f(a)-(x-a)f'(a)-\frac{(x-a)^2}{2}f''(a)-\dots-\frac{(x-a)^n}{n!}f^{(n)}(a).$$ We prove that for all $a,x\in(b,d)$ $$R_n(x)=\frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(c)$$ for some $c$ between $a$ and $x$.

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Proof:

By induction on $n$. For $n=0$ this is same as Mean Value Theorem. For $n\in\mathbb{N}$, notice $$R'_n(x)=f'(x)-f'(a)-(x-a)f''(a)-\dots-\frac{(x-a)^{n-1}}{(n-1)!}f^{(n)}(a).$$ Applying inductive hypothesis on $f'$, we get $$R'_n(x)=\frac{(x-a)^n}{n!}f^{(n+1)}\left(c(x)\right)$$ for some $c(x)$ between $a$ and $x$. So $$R_n(x)=R_n(a)+\int_a^x\frac{(t-a)^{n}}{n!}f^{(n+1)}\left(c(t)\right)dt$$ where $R_n(a)=0$. By Weighted Mean Value Theorem, which requires that $f^{(n+1)}\left(c(t)\right)$ is continuous and $(t-a)^n/n!$ does not change sign between $a$ and $x$, we have $$R_n(x)=f^{(n+1)}(c)\int_a^x\frac{(t-a)^{n}}{n!}dt=\frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(c)$$ for some $c$ between $a$ and $x$.

Caveat: we used the fact without proof that $c(x)$ is continuous over $[b,d]$ for $R'_n(x)$.