Prove that $\prod_{i=1}^n(1+x_i)\leq \sum_{i=0}^n\frac{S^i}{i!}$, where $x_i\in\mathbb{R^+}$. 
Let $x_1$, $x_2$, $\ldots$, $x_n$ be positive real numbers, and let $$S=x_1+x_2+\cdots+x_n.$$ Prove that $$(1+x_1)(1+x_2)\cdots(1+x_n)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots+\frac{S^n}{n!}.$$

My first thought is about using induction. For $n=1$, $LHS=1+x_1\leq1+S=RHS$.
Now I suppose that this inequality holds true for $n=k$; that is, $$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$
And here I am stuck. I can't really see how I can find the relation between when $n=k$ and $n=k+1$.
Perhaps induction may not be the way?  Any hints or suggestions will be much appreciated. Thanks.

Show that
$$ i! \sum_{1 \leq a_1 < a_2 \ldots < a_i \leq n } x_{a_1} x_{a_2} \ldots x_{a_i}  \leq S^i.$$
This is almost immediately obvious by expanding the RHS and noticing that we have a ton of extra terms.
In fact, we have strict inequality when $n \geq 2$.
Then, sum from $i = 0$ to $n$ to obtain the desired inequality.
