Show $(\frac{x^n}{n!})'=\frac{x^{n-1}}{(n-1)!}$

Question

Let $n\in\mathbb{N}_0$ and $x\in\mathbb{C}$ with $x^{[n]}=\frac{x^n}{n!}$ and $x^{[-1]}=0$. Show for $x,y\in\mathbb{C}$ that $(x^{[n]})'=x^{[n-1]}$.

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Our lecture has not introduced integrals yet, let alone the gamma function, we have only so far covered differential calculus up to the Taylor series.

I know that $\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$, but surely the derivative of $e^x$ is $e^x$ itself again and thus not what we're looking here for…right? Can anybody suggest a workable approach to me? I'm quite stumped right now, even though this exercise is apparently intended to be not that hard.

Answer 1

Differentiating with respect to $x$ means you keep the other constants as a variable. Therefore, ignore the $n!$ in the denominator.$$f(x)=\frac {x^n}{n!}\qquad\implies\qquad f'(x)=\frac {nx^{n-1}}{n!}$$Expand the factorial as the product of integers and simplify$$\begin{align*}f'(x) & =\frac {nx^{n-1}}{n(n-1)(n-2)\cdots1}\\ & =\frac {x^{n-1}}{(n-1)(n-2)\cdots1}\\ & =\frac {x^{n-1}}{(n-1)!}\end{align*}$$

Answer 2

\begin{align} \frac{d}{dx} \left( \frac{x^n}{n!}\right) &= \frac1{n!} \left[\frac{d}{dx}x^{n}\right] \end{align}

Simplify the term in the square bracket and then use the property of factorial to get your desired answer.

Answer 3

Differentiating with respect to $x$ gives $$\frac{d}{dx}\Big(\frac{x^n}{n!}\Big)=\frac{1}{n!}\frac{d(x^n)}{dx}=\frac{1}{n!}nx^{n-1}=\frac{1}{n(n-1)(n-2)...(1)}nx^{n-1}=\frac{x^{n-1}}{(n-1)!}.$$