Problem Let $p_{n-1}$ be a polynomial of degree $\leq(n-1)$ that interpolates the function $f(x)=e^{x}$ at any set of $n$ nodes in the interval $[-1,1]$. Prove that the error obeys this inequality on $[-1.1]$, $$p_{n-1}(x)-f(x)\leq \frac{2^{n}e}{n!} \forall x[-1,1] $$
Solution: From interpolation error formula for $n$ nodes, we have $$p_{n-1}(x)-f(x)\leq \frac{(x-x_{1})(x-x_{2})(x-x_{3})......(x-x_{n})}{n!}f^{n}(c)$$ Where $c\in[-1,1]$. Here, $f(x)=e^{x}$. Therefore, $|f^{n}(c)|\leq e^{1}$. And, because the interval is $[-1,1]$. The maximum difference froma a specified point can be $2$. Thus , $$p_{n-1}(x)-f(x)\leq \frac{2^{n}e}{n!} \forall x[-1,1] $$
Is my solution correct? And can anyone give a more articulate solution similar to mine. As my solution looks way too casual.