Proving the result $$(x+n)^n-\binom{n}{1}(x+n-1)^{n}+\binom{n}{2}(x+n-2)^{n}\cdots(-1)^nx^n=n!$$

Try: $$\bigg(x^n+\binom{n}{1}x^{n-1}n+\binom{n}{2}x^{n-2}n^2+\cdots n^n\bigg)-\binom{n}{1}\bigg(x^n+\binom{n}{1}x^{n-1}(n-1)\cdots\bigg)+\cdots +(-1)^n\binom{n}{n}x^n$$

Could some help me to solve it, thanks


2 Answers 2


It's convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write e.g. $$n![z^n]e^{kz}=k^n$$

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n}{k}(x+n-k)^n(-1)^k}\\ &=\sum_{k=0}^n\binom{n}{k}(x+k)^n(-1)^{n-k}\tag{1}\\ &=\sum_{k=0}^n\binom{n}{k}n![z^n]e^{(x+k)z}(-1)^{n-k}\tag{2}\\ &=n![z^n]e^{xz}\sum_{k=0}^n\binom{n}{k}\left(e^z\right)^k(-1)^{n-k}\tag{3}\\ &=n![z^n]e^{xz}\left(e^z-1\right)^n\tag{4}\\ &=n![z^n]e^{xz}\left(z+\frac{z^2}{2!}+\cdots\right)^n\tag{5}\\ &\color{blue}{=n!} \end{align*} and the claim follows.


  • In (1) we change the order of summation by setting $k \rightarrow n-k$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

  • In (2) we apply the coefficient of operator.

  • In (3) we use the linearity of the coefficient of operator and do some rearrangements in order to apply the binomial theorem.

  • In (4) we apply the binomial theorem.

  • In (5) we do the series expansion of $e^{z}$ and see the smallest exponent of $z$ in $(e^z-1)^n$ is $\color{blue}{n}$. So, we will only use the constant $\color{blue}{1}$ in the series expansion of $e^{xz}=\color{blue}{1}+xz+\frac{(xz)^2}{2}+\cdots$ in order to obtain the coefficient of $z^n$, resulting in $$[z^n]e^{xz}\left(e^z-1\right)^n=1$$


I apply a one-to-one continuation of Jack D'Aurizio's answer to your (-;) question in here.

Let $p(x)$ be a polynomial with degree $d\geq 1$, and let $\delta$ be the (difference) operator mapping $p(x)$ into $p(x)-p(x-1)$. Then $(\delta p)(x)$ is a polynomial with degree $d-1$. We have that

$$(\delta^n p)(x) = \sum_{k=0}^{n}(-1)^k \binom{n}{k} p(x-k) \tag{1}$$

The sum you ask for is the RHS of $(1)$ in the case $p(x)=(x+n)^{n}$. So the original polynomial had degree $d=n$, hence $(1)$ has degree $n-n=0$ in $(x+n)$ which is to say that the result is the sum of all terms with power zero of $(x+n)$. This is given as

$$ \sum_{k=0}^{n}(-1)^k \binom{n}{k} (-k)^n= (-1)^n \sum_{k=0}^{n}\binom{n}{k} (-1)^k \frac{\partial^n}{(\partial a)^n}\exp (ak) |_{a=0} \\ = (-1)^n \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} \sum_{k=0}^{n}\binom{n}{k} (-e^a)^k = \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} (e^a-1)^n \\ = \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} a^n(1 +a/2 + \dots )^n = \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} a^n = n! $$

where in the last line, after expansion of the exponential, the fact was used that only terms with $a^n$ survive the operator $\lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n}$.

This proves the claim. $\qquad \Box$


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