Showing $(n)^{\overline{k}}=(n+k-1)^{\underline{k}}=(-1)^k\cdot(-n)^{\underline{k}}$, for rising and falling factorials

Question

I have to prove equality with falling factorials $$(n)^{\underline{k}} = n(n-1)\cdots(n-k+1) $$ and rising factorials $$(n)^{\overline{k}}=n(n+1)\cdots(n+k-1)$$

Show: $$(n)^{\overline{k}}=(n+k-1)^{\underline{k}}= (-1)^k \cdot(-n)^{\underline{k}}$$

Can anyone please give me a hint? I tried so many ways but it got nowhere...my tries:

$$(n+k-1)^{\underline{k}}=(n+k-1)(n+k-2)...(n+k)$$ $$(-n)^{\underline{k}}=-n(-n-1)...(-n-k+1)$$

Answer 1

It is also convenient to use the product symbol to explicitly see the lower and upper limit. We have \begin{align*} \color{blue}{n^{\overline{k}}}&=n(n+1)\cdots(n+(k-1) \color{blue}{=\prod_{j=0}^{k-1}(n+j)}\tag{1}\\ \color{blue}{n^{\underline{k}}}&=n(n-1)\cdots(n-(k-1)) \color{blue}{=\prod_{j=0}^{k-1}(n-j)}\tag{2} \end{align*}

We obtain \begin{align*} \color{blue}{n^{\overline{k}}} &=\prod_{j=0}^{k-1}(n+j)\tag{3}\\ &=\prod_{j=0}^{k-1}(n+(k-1-j))\tag{4}\\ &\,\,\color{blue}{=(n+k-1)^{\underline{k}}}\tag{5}\\ \\ \color{blue}{n^{\overline{k}}} &=\prod_{j=0}^{k-1}(n+j)\tag{3}\\ &=(-1)^k\prod_{j=0}^{k-1}(-n-j)\tag{6}\\ &\,\,\color{blue}{=(-n)^{\underline{k}}}\tag{7} \end{align*} and the claim follows.

Comment:

• In (3) we use the product representation (1).

• In (4) we change the order of multiplication by replacing the index $j$ with $k-1-j$. \begin{align*} \prod_{j=0}^{k-1}(n+j)&=(n+0)(n+1)\cdots (n+k-2)(n+k-1)\\ &=(n+(k-1))(n+(k-2))\cdots(n+1)(n+0)\\ &=\prod_{j=0}^{k-1}(n+(k-1-j)) \end{align*}

• In (5) we use (2) with $n$ substituted by $n+k-1$.

• In (6) we factor out $(-1)$ $k$ times.

• In (7) we use (2) with $n$ substituted by $-n$.

Answer 2

Hints.

In the first line you made an error. It should be (please check!): $$(n+k-1)^{\underline{k}}=(n+k-1)(n+k-2)\cdots\color{red}n$$ Now recall that multiplication is commutative and revert the factors!

In the second line everything is correct. Now multiply every factor by $-1$. To keep the correct sign of the whole expression multiply it by another $-1$ for each factor.