Taylor's Lagrange remainder follows from repeated Rolle steps

Fix $x\mathrm{\ne }c$ and set $A=(f(x)-P_n(x))\mathrm{/}(x-c{)}^{n+1}$ and $g(t)=f(t)-P_n(t)-A(t-c{)}^{n+1}$ on the interval between $c$ and $x$. Then $g(c)=g(x)=0$ and $g^{(j)}(c)=0$ for each $1\le j\le n$, because the Taylor polynomial matches those derivatives at $c$ by [L2] and the last term has a zero of order $n+1$.

By [L1] applied to $g$ on the interval with endpoints $c,x$, there is a point $z_1$ strictly between them with $g^{\mathrm{\prime }}(z_1)=0$. If $n\ge 1$, also $g^{\mathrm{\prime }}(c)=0$, so [L1] applied to $g^{\mathrm{\prime }}$ between $c$ and $z_1$ gives $z_2$ between them with $g^{\mathrm{\prime }\mathrm{\prime }}(z_2)=0$. Inductively, for each $j\le n$ the zeros $g^{(j)}(c)=0$ and $g^{(j)}(z_j)=0$ give a point $z_{j+1}$ between them with $g^{(j+1)}(z_{j+1})=0$. The last point is the required $\xi$; when $n=0$, take $\xi =z_1$.

Since $P_n^{(n+1)}=0$ and the $(n+1)$st derivative of $A(t-c{)}^{n+1}$ is $A(n+1)!$, the equation $g^{(n+1)}(\xi )=0$ gives $A=f^{(n+1)}(\xi )\mathrm{/}(n+1)!$ by [L2]. Substituting the definition of $A$ proves the formula. At $x=c$, $P_n(c)=f(c)$ directly.