Let $a_i, 1 \leq i \leq n$ be non-negative real numbers. Let $S$ denote their sum. To prove $$\prod_{k=1}^n(1+a_k) \leq 1 + \frac{S}{1}+\frac{S^2}{2!} + \ldots + \frac{S^n}{n!}$$
Let $S = \sum_{k=1}^n a_k$. Using the inequality $\log(1 + x) \le x$, and the fact that $e^x$ is monotonically increasing, we have $$ \prod_{k=1}^n (1 + a_k) = e^{\log\prod_{k=1}^n(1 +a_k)} = e^{\sum_{k=1}^n \log(1 +a_k)} $$ $$ \le e^{\sum_{k=1}^n a_k} = e^S $$
I am not getting the tighter bound. How to look for it?
Let $$S_k=\frac 1 {n \choose k} \displaystyle \sum_{1\leq i_{1}<\cdots <i_{k}\leq n} a_{i_1}a_{i_2}\cdots a_{i_k}.$$ Then $\displaystyle \prod_{k=1}^n(1+a_k)=1+\sum_{k=1}^n {n \choose k} S_k$. Using Maclaurin's inequality, $$\prod_{k=1}^n(1+a_k) \le 1+ \sum_{k=1}^n {n \choose k} S_1^k = 1 + \sum_{k=1}^n \frac {n!}{k!(n-k)!} \frac {S^k}{n^k} \\ = 1 + \sum_{k=1}^n \frac {n(n-1)\dots(n-k+1)}{n^k} \frac {S^k}{k!} \le 1 + \sum_{k=1}^n \frac {S^k}{k!}.$$
We can even do without Maclaurin's inequality, using only AM–GM: $$\sqrt[n]{\prod_{k=1}^n (1+a_k)} \le \frac {\displaystyle \sum_{k=1}^n (1+a_k)} n = 1 + \frac S n.$$ From this, we have: $$\prod_{k=1}^n (1+a_k) \le \left(1+\frac S n\right)^n = 1+ \sum_{k=1}^n \frac {n!}{k!(n-k)!} \frac {S^k}{n^k}.$$The rest is as above. Note that is it sufficient that $a_k \ge -1, \forall k$ and $S = \sum_{k=1}^n a_k \ge 0$.