Prove that if $f:\mathbb R^n \to \mathbb R$ that $f\in C^{k+1}$ in the closed ball $B$ then $$\sum_{|\beta|=k+1} \frac{|\beta|}{\beta!} \int_0^1 (1-t)^{|\beta|-1}D^\beta f \big(t \boldsymbol{x} \big) \, dt(\boldsymbol{x})^\beta=\sum_{|\beta|=k+1} R_\beta(\boldsymbol{x})(\boldsymbol{x})^\beta$$ for some $R_\beta \in C^{\infty}$
I know that it is the fact from here: Taylor's theorem about the formula for the remainder, but I don't have any idea how to prove it...
For clarity: in this task we assume that the remainder has the form $\sum_{|\beta|=k+1} \frac{|\beta|}{\beta!} \int_0^1 (1-t)^{|\beta|-1}D^\beta f \big(t \boldsymbol{x} \big) \, dt(\boldsymbol{x})^\beta$ and we have to prove that it is equal $\sum_{|\beta|=k+1} R_\beta(\boldsymbol{x})(\boldsymbol{x})^\beta$ for some $R_\beta \in C^{\infty}$ (not the other way around).
Is there anyone who could help me? I will appreciate every hint!
First, you need to assume that $f$ is smooth. If you set
$$ R_{\beta}(x) = \frac{|\beta|}{\beta!} \int_0^1 (1-t)^{|\beta| - 1} (D^{\beta} f)(tx) \, dt $$
and differentiate $R_{\beta}$ under the integral sign, you get
$$ (D^{\alpha}R_{\beta})(x) = \frac{|\beta|}{\beta!} \int_0^1 (1-t)^{|\beta| -1} t^{|\alpha|} (D^{\beta + \alpha}f)(tx) \, dt $$
so all the partial derivatives of $R_{\beta}$ of all orders exist and $R_{\beta}$ is smooth. If you want to argue more formally, prove by induction that all partial derivatives of $R_{\beta}$ of order less than $k$ exist and are given by the formula above and use differentiation under the integral sign to deduce that all partial derivatives of $R_{\beta}$ of order less than $k + 1$ exist and are given by the same formula.