If $S = x_1 + x_2 + .. + x_n$, Prove that $ (1+x_1)(1+x_2)..(1+x_n) \le 1 + S + \frac{S^2}{2!} + .. + \frac{S^n}{n!}$

By $AM-GM$ we have $$(1+x_1)(\dots)(1+x_n) \le (1+ \frac{1}{n}S)^n = \sum_{k=0}^n a_kS^k $$ where $$a_k = \frac{n!}{n^k (n-k)!} \frac{1}{k!}\le \frac{1}{k!}.$$

Hint:

$$\prod_{k=1}^n (1+x_k) \leqslant (1 +S/n)^n = 1 + n \frac{S}{n} + \frac{n(n-1)}{2!} \frac{S^2}{n^2} + \ldots + \frac{n(n-1)\ldots 2\cdot 1}{n!}\frac{S^n}{n^n} \\= 1 + S + \frac{1-1/n}{2!}S^2 + \ldots + \frac{(1-1/n)(1-2/n)\ldots (1 - (n-1)/n)}{n!}S^n $$

I use this fact :$$(x_1+x_2+x_3...+x_n)^n=\sum_{k_1+k_2+...k_j=n} \dfrac{n!}{k_1!k_2!...}x_1^{k_1}x_2^{k_2}...x_n^{k_j}$$ $$1 \leq 1\\(x_1+x_2...+x_n)\leq S\\(x_1x_2+x_1x_3+...+x_{n-1}x_n) \leq \frac12S^2=\frac12(2(x_1x_2+x_1x_3+...)+(x_1^2+x_2^2+...x_n^2))\\(x_1 x_2x_3+x_1x_2x_4+...)\leq\frac{1}{3!}S^3=\frac{1}{6}(\color{red} {\frac{3!}{1!1!1!}(x_1x_2x_3+x_1x_2x_4+...)}+\frac{3!}{2!1!}(x_1^2x_2+x_2^2x_1+...)+\frac{3!}{3!}(x_1^3+x_2^3+...+x_n^3))\\$$

$$(x_1 x_2x_3x_4+x_1x_2x_3x_5+...)\leq\frac{1}{4!}S^4=\frac{1}{24}(\color{red} {\frac{34!}{1!1!1!1!}(x_1x_2x_3x_4+x_1x_2x_3x_5+...)}+\frac{4!}{3!1!}(x_1^3x_2+x_2^3x_1+...)+\frac{4!}{2!2!}(x_1^2x_2^2+...)+\frac{4!}{4!}(x_1^4+x_2^4+...+x_n^4))$$ and so on

now we have $$1\leq1\\+(x_1+x_2+...x_n)\leq\frac{1}{2!}S^2\\ +(x_1x_2+x_1x_3+...)\leq\frac{1}{3!}S^3\\ +(x_1x_2x_3+x_1x_2x_4+...)\leq\frac{1}{4!}S^4\\+...\\ +(x_1x_2x_3...x_n)\leq\frac{1}{n!}S^n\\ 1+(x_1+x_2+...x_n)+(x_1x_2+x_1x_3+...)+...(x_1x_2...x_n)\leq 1+s+\frac{S^2}{2}+...+\frac{S^n}{n!}\\$$ and $$(1+x_1)(1+x_2)...(1+x_n)=1+(x_1+x_2+...x_n)+(x_1x_2+x_1x_3+...)+...(x_1x_2...x_n)$$proof is complete now