See the pattern in the following product involving e

Question

I've the following pattern, how do I write it in product form:

 1. $$e-\text{k}\cdot\text{x}_{\text{k}-1}$$  2. $$e-\text{k}\cdot\left(e-\left(\text{k}-1\right)\cdot\text{x}_{\text{k}-2}\right)$$  3. $$e-\text{k}\cdot\left(e-\left(\text{k}-1\right)\cdot\left(e-\left(\text{k}-2\right)\cdot\text{x}_{\text{k}-3}\right)\right)$$  4. $$e-\text{k}\cdot\left(e-\left(\text{k}-1\right)\cdot\left(e-\left(\text{k}-2\right)\cdot\left(e-\left(\text{k}-3\right)\cdot\text{x}_{\text{k}-4}\right)\right)\right)$$  5. $$e-\text{k}\cdot\left(e-\left(\text{k}-1\right)\cdot\left(e-\left(\text{k}-2\right)\cdot\left(e-\left(\text{k}-3\right)\cdot\left(e-\left(\text{k}-4\right)\cdot\text{x}_{\text{k}-5}\right)\right)\right)\right)$$

.... And so on

Where $e\approx2.718281828459045$

Answer 1

Notice that we have

$$e-k(\underbrace{e-(k-1)x_{k-2}}_{x_{k-1}})$$

So, we can try some induction, if we can assume the this will take. Looking at the fourth line, we find

$$e-k(e-(k-1)(e-(k-2)(e-(k-3)x_{k-4})))\\=e-ke+k(k-1)e-k(k-1)(k-2)e+k(k-1)(k-2)(k-3)x_{k-4}\\=\frac{k!(-1)^4}{(k-4)!}x_{k-4}+e\sum_{p=0}^3\frac{k!(-1)^p}{(k-p)!}$$

So, in general, we should have

$$n^{\text{th}}\text{ line}=\frac{k!(-1)^n}{(k-n)!}x_{k-n}+e\sum_{p=0}^{n-1}\frac{k!(-1)^p}{(k-p)!}$$

Whereupon we find

$$(n+1)^{\text{th}}\text{ line}=\frac{k!(-1)^n}{(k-n)!}(e-(k-n)x_{k-n-1})+e\sum_{p=0}^{n-1}\frac{k!(-1)^p}{(k-p)!}$$

$$\begin{align}&=e\frac{k!(-1)^n}{(k-n)!}+(-1)\frac{k!(-1)^n}{(k-n)!}(k-n)x_{k-n-1}+e\sum_{p=0}^{n-1}\frac{k!(-1)^p}{(k-p)!}\\&=\frac{k!(-1)^{n+1}}{(k-n-1)!}x_{k-n-1}+e\sum_{p=0}^n\frac{k!(-1)^p}{(k-p)!}\end{align}$$

So, by induction, since this is true for $e-kx_{k-1}$, it holds true for all products of this type.

Answer 2

With $x_k=e-k·x_{k-1}$ you can define a second sequence $y_k=(-1)^k\frac{x_k}{k!}$ which has the iteration $$ y_k=(-1)^k\frac{e-k·x_{k-1}}{k!}=(-1)^k\frac{e}{k!}+(-1)^{k-1}\frac{x_{k-1}}{(k-1)!}=e\frac{(-1)^k}{k!}+y_{k-1} $$ which converges to $$ \lim_{n\to\infty} y_n=y_0-e+e\sum_{k=0}^\infty \frac{(-1)^k}{k!}=y_0-e+e·\exp(-1)=y_0-e+1 $$