Proof by induction for nth derivative Show the following hold by induction:
$$\frac {d^n}{dx^n}\frac {e^x - 1}{x} = (-1)^n \frac{n!}{x^{n+1}} \left( e^x \left(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}\right) - 1 \right)$$
Proof.
It's not hard to show the base case hold.
For inductive step, we can also write this as:
$$\frac {d^n}{dx^n}\frac {e^x - 1}{x} = (-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!} - (-1)^n \frac{n!}{x^{n+1}} $$
Take derivative on both side:
$$\frac {d^{n+1}}{dx^{n+1}}\frac {e^x - 1}{x} = \frac {d}{dx}(-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!} - \frac {d}{dx}(-1)^n \frac{n!}{x^{n+1}} $$
$$\frac {d^{n+1}}{dx^{n+1}}\frac {e^x - 1}{x} = \underbrace{\frac d{dx}(-1)^n\frac{n!}{x^{n+1}}e^x\sum_{k=0}^n(-1)^k\frac{x^k}{k!}}_? - \underbrace{(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}}_{hold} $$
Therefore, my question is for the first part, how do I show the following hold:
$$ \frac d{dx}(-1)^n\frac{n!}{x^{n+1}}e^x\sum_{k=0}^n(-1)^k\frac{x^k}{k!} 
= (-1)^{n+1}\frac{(n+1)!}{x^{n+2}} e^x \sum_{k=0}^{n+1} (-1)^{k} \frac{x^k}{k!}$$
 A: It looks like you need to use the product rule here:
$$\frac {d}{dx}((-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!})$$
$$=(-1)^n \frac{d}{dx}(\frac{n!}{x^{n+1}}) e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} \frac{d}{dx}(e^x) \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} e^x \frac{d}{dx}(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!})$$
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}}e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=1}^{n} (-1)^{k} \frac{x^{k-1}}{(k-1)!}$$
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}}e^x \sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}} e^x \sum_{k=0}^{n-1} (-1)^{k+1} \frac{x^{k}}{k!}$$
Now notice that the last two terms cancel each other out except for the extra $n$th term in the middle sum. So this becomes
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^n \frac{n!}{x^{n+1}}e^x (-1)^{n} \frac{x^n}{n!}$$
$$=(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}
+(-1)^{n+1} \frac{(n+1)!}{x^{n+2}}e^x (-1)^{n+1} \frac{x^{n+1}}{(n+1)!}$$
Here I just multiplied by $(-1)(-1)\frac{(n+1)}{(n+1)}\frac{x}{x}=1$ in the second term. Finally, we use the distributive property to finish it off.
$$(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\left(\sum_{k=0}^{n} (-1)^{k} \frac{x^k}{k!}+(-1)^{n+1} \frac{x^{n+1}}{(n+1)!}\right)$$
$$(-1)^{n+1}\frac{(n+1)!}{x^{n+2}}e^x\sum_{k=0}^{n+1} (-1)^{k} \frac{x^k}{k!}$$
