Show that $(x_1+1)(x_2+1)...(x_n+1)=1+x_1 +x_2+...x_n+$ "other terms". 
Show that $(x_1+1)(x_2+1)...(x_n+1)=1+x_1 +x_2+...x_n+$ "other terms".


How can the following be expanded ?
$$(x_1+1)(x_2+1)...(x_n+1)$$
I have a problem where I have to expand and show that I'll have a form of $1+x_1 +x_2+...x_n+$all the other terms. I don't have a math background so I'm trying to figure out an appropriate way to show this. 
 A: Rewrite it as follows:
$$(1+x_1)(1+x_2)\dots(1+x_n)$$
Then, foil, starting from the front and working our ways through.
$$(1+x_1+x_2+x_1x_2)(1+x_3)\dots(1+x_n)$$
However, you are only interested in the terms $1$ and anything that isn't a product of $2$ of the numbers, so...
$$(\color{red}{1+x_1+x_2}+\color{blue}{x_1x_2})(1+x_3)\dots(1+x_n)\implies(1+x_1+x_2)(1+x_3)\dots(1+x_n)$$
We're really only interested in the red terms.  Anyways, we foil again and drop everything we don't want,
$$(\color{green}{1+x_1+x_2+x_3})(1+x_4)\dots(1+x_n)$$
$$(\color{green}{1+x_1+x_2+x_3+x_4})(1+x_5)\dots(1+x_n)\\\vdots\\\color{green}{1+x_1+x_2+\dots+x_n}$$
So, the original product is equal to this plus everything we dropped,
$$1+x_1+x_2+\dots+x_n+\text{other stuff}$$
If you repeat this process to include everything that is $1$, $x_k$, or a product of two terms, you'll get the above plus more information.
Repeatedly applying this, there is a nice general form...
$$(1+x_1)(1+x_2)\dots(1+x_n) =$$
$$\begin{align}
\qquad& 1\\
& +x_1+x_2+\dots+x_n \\
& +x_1(x_2+\dots+x_n)+x_2(x_3+\dots+x_n)+\dots+x_{n-1}(x_n) \\
& +x_1(x_2(x_3+\dots+x_n)+x_3(x_4+\dots+x_n))+\dots)+x_2(\dots)+\dots+x_{n-2}(x_{n-1}(x_n)) \\
& +\dots \\
& \text{more and more parenthesis per line}
\end{align}$$

An amazing use of this technique was exploited by Euler in solving the Basel problem (expansion of the product step).
A: From a layman's stand point, notice that each expression in parentheses has 2 terms.  And there are $n$ expressions.  Let's work slowly through a few examples.
If $n=2$, we have $(x_1+1)(x_2+1)$.  By an elementary property called distribution done twice we can show the following...
$$(x_1+1)(x_2+1)=(x_1+1)\cdot x_2+(x_1+1)\cdot 1=x_1x_2 + x_2 +x_1+1$$
Notice there are $4=2^2$ terms; a 2 term expression multiplied by a 2 term expression yields a 4 term expression.  Knowing now what we know, consider the next step, $n=3$:
$$(x_1+1)(x_2+1)(x_3+1)=[(x_1+1)\cdot x_2+(x_1+1)\cdot 1](x_3+1)$$
$$=(x_1x_2 + x_2 +x_1+1)(x_3+1)$$
and by distribution again we get
$$x_1 x_2 x_3+x_2 x_3+x_1 x_3+x_3+x_1 x_2+ x_2+x_1 +1$$
A nd now there are $8=2^3$ terms; a ... continue to determine the rest and hopoefully you see the pattern.  For our $n=3$ case, there is 1 term which is the product of all the $1$'s.  There are 3 terms which is the sum of the single disjoint $x$ terms.  There 3 terms which are the product of 2 different $x's$ of different subscripts, and there is 1 term which is the product of all 3 subscripted terms.  THis can be generalized.  Look up binomial expansion or the binomial theorem to see more information.
A: Since this turned into a guessing game of what the question really asks, here is another take on it.
Let $P(x)=(x+x_1)(x+x_2)\cdots(x+x_n)$ be the polynomial with roots $-x_1, -x_2,\cdots,-x_n$.
$P(x)$ is a monic $n^{th}$ degree polynomial, so it can be written as
$P(x)=x^n + a_{n-1}x^{n-1}+\cdots+a_0$.
Then $(x_1+1)(x_2+1)\cdots(x_n+1)=P(1)=1 + a_{n-1} + \cdots + a_0$.
But $a_{n-1}=x_1+x_2+\cdots+x_n$ by Vieta's formulas so $$(x_1+1)(x_2+1)\cdots(x_n+1)=1 + (x_1+x_2+\cdots+x_n) + \text{<other terms>}$$
where $\text{<other terms>} = a_{n-2} + \cdots +a_0$.
A: Yes you can prove by induction that
$$(x_1 + 1)(x_2 + 1) \ldots (x_n+1) = 1 +\sum_{j =1}^n\sum_{k_1=1}^n\underbrace{\overbrace{\sum_{k_2=k_1+1}^n\sum_{k_3=k_2+1}^n\dots\sum_{k_j=k_{j-1}+1}^n}}_{\huge\underbrace{\dots\sum_{k_p=k_{p-1}+1}^n\dots}_{\large\text{magnified}}}^{\large j-1\text{ sums}} x_{k_1}\ldots x_{k_j}$$
Consider the case $j=1$.
