Prove the equality : n!=Σ_(k=0) ^n (−1)^k ((n),(k) )(n−k)^n


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Question Number 122624 by Ar Brandon last updated on 18/Nov/20

$Prove the equality :$ $n! = \underset{k = 0}{\sum^{n}} {(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix}) {(n - k)}^{n}$
	Answered by mindispower last updated on 18/Nov/20

	$l e t f (x, y) = e^{y} {(e^{y} - x)}^{n}$ $f (x, y) = e^{y} [\underset{k = 0}{\sum^{n}} {(- x)}^{k} C_{k}^{n} e^{(n - k) y}]$ $= \underset{k = 0}{\sum^{n}} e^{(n + 1 - k) y} {(- 1)}^{k} x^{k} C_{k}^{n}$ $f (1, y) = \underset{k = 0}{\sum^{n}} e^{(n + 1 - k) y} {(- 1)}^{k} C_{n}^{k}$ $l e t \partial_{y}^{k} f (1, y) b e e k d e r i v a t e f o v e r y$ $w e w a n t \partial_{y}^{n} f (1, 0)$ $\partial_{y}^{n} f (1, y) = \sum_{k ⩽ n} {(n + 1 - k)}^{n} e^{(n + 1 - k) y} {(- 1)}^{k} C_{n}^{k}$ $\partial_{y}^{n} f (1, 0) = \sum_{k ⩽ n} {(n + 1 - k)}^{n} {(- 1)}^{k} C_{n}^{k}$ $f (1, y) = e^{y} {(e^{y} - 1)}^{n}$ $= e^{y} {(e^{\frac{y}{2}} (e^{\frac{y}{2}} - e^{- \frac{y}{2}}))}^{n}$ $= e^{(1 + \frac{n}{2}) y} 2^{n} {s h}^{n} (\frac{y}{2}) = g (\frac{y}{2}) = f (1, y)$ $l e t \frac{y}{2} = z$ $g (z) = h (y), \frac{\partial h}{\partial y} = \frac{\partial h}{\partial z} . \frac{\partial z}{\partial y} = \frac{\partial g}{\partial z} . \frac{1}{2}$ $\Rightarrow \partial^{n} h (y) = \frac{1}{2^{n}} . \frac{\partial^{n} g}{\partial z^{n}}$ $\Rightarrow \partial_{y}^{n} f (1, y) ∣_{y = 0} = \partial_{z}^{n} e^{(2 + n) z} {s h}^{n} (z) ∣_{z = 0}$ $w e s h o w t h a t \forall k < n \partial_{x}^{k} {s h}^{n} (x) ∣_{x = 0} = 0 . . E$ $\partial_{x} {s h}^{n} (x) = n c h (x) {s h}^{n - 1} (x)$ $w e k n o w s h (0) = 0$ $a l l {s h}^{i} (x) ∣ x = 0 g i v e z e r o o n l y i = 0$ $b u t d e r i v a t e k < n t h e l e a s t p o w e r o f s h i s n - k ⩾ 1$ $\Rightarrow w e h a v e a l l p o w e r o f {s h}^{m} (x), m \in [n - k, n]$ $= 0$ $\Rightarrow \partial_{z}^{n} e^{2 + n} {s h}^{n} (z) = Σ C_{n}^{k} {(e^{(2 + n) z})}^{(n - k)} {s h}^{n} {(z)}^{(k)}$ $b y p r o p o s i t i o n ? E ?$ $w e n o w t h a t$ $t h e o n l y f a c t e u r w e w a n t$ ${i s C}_{n}^{n} e^{(2 + n) z} {({s h}^{n} (z))}^{(n)} ∣_{z = 0}$ ${s h}^{n} {(z)}^{(n)} = {(n c h (z) {s h}^{n - 1} (z))}^{n - 1} . . a p p l y E a g a i n e$ $w e w a n t n c h (z) {({s h}^{n - 1} (z))}^{n - 1}$ $= n (c h (z) (n - 1) c h (z) {({s h}^{n - 2} (z))}^{n - 2} . . . . .$ $= n (n - 1) (n - 2) . . . . . . . . (n - (n - 1)) {({s h}^{n - (n - 1} (z))}^{(n - (n - 1))}$ $= (n)! . {(s h (z))}^{^{'}} ∣_{z = 0}$ $= n! c h (0) = n!$
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