Show that $\vert\int_{-1}^1 \omega(t) dt \vert \leq 2^n \int_{-1}^1\vert \omega^{(n)}(t)\vert dt$

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I am stuck with the following problem:

With $\omega: [-1,1]\rightarrow \mathbb{R}$, $\omega\in C^n(-1,1)$. Suppose that $\omega$ has a finite number of zeroes $t_1<t_2<\cdots <t_n$ (i.e. $\omega(t_i)=0,\forall i$) on $[-1,1]$. Prove that $$\left\vert\int_{-1}^1 \omega(t) dt \right\vert \leq 2^n \int_{-1}^1\vert \omega^{(n)}(t)\vert dt$$

I think I should show it inductively, but I can figure out how to do it. If someone could give me some hints that would be greatly appreciated.

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Applying Rolle's theorem an awful lot of times, you will find that all derivatives of $\omega$ of orders $\le n-1$ have at least one zero in $[-1,1]$. This allows you to run the following, outrageously wasteful, chain of inequalities for $k=0,1,\dots, n-1 $:

$$\int_{-1}^1 |\omega^{(k)}(x)|\,dx \le 2 \sup_{[-1,1]} |\omega^{(k)}(x)| \le 2\int_{-1}^1 |\omega^{(k+1)}(x)|\,dx$$ where the last step uses the fundamental theorem of calculus: $\omega^{(k)}(x)$ is given by the integral of $\omega^{(k+1)} $ from some zero of $\omega^{(k)}$ to $x$.