Boutin's Identity I would like to speak about the identity that I found on the website of Tito Piezas this is it :

$$\sum_{2^{n-1}}\pm(\pm x_1\pm x_2\pm\cdots\pm x_n)^n=n!2^{n-1}x_1\cdots x_n$$

Where the exterior sign being the product of the $n$ interior signs and $x_i$ are reals.
My try :
We derivate with respect to $x_1$ we get :
$$\sum_{2^{n-1}}\pm n(\pm x_1\pm x_2\pm\cdots\pm x_n)^{n-1}=n!2^{n-1}x_2\cdots x_n$$
We do the same thing for $x_2,\cdots ,x_{n-1}$ we get :
$$\sum_{2^{n-1}}\pm (\pm x_1\pm x_2\pm\cdots\pm x_n)=2^{n-1} x_n$$
Wich is obvious .
Thanks a lot for your contributions.
 A: I think a better notation for the sum will help you solve the problem. The Boutin identity can be stated as 
$$\sum_{\bar{\sigma}\in S} (-1)^{\sum_{l=1}^{k-1}\sigma_l}\left(x_1+(-1)^{\sigma_1}x_2+(-1)^{\sigma_2}x_3+\ldots+(-1)^{\sigma_{k-1}}x_k\right)^k = k! 2^{k-1} x_1 x_2 \ldots x_k $$
where $S = \{0,1\}^{\times (k-1)}$ and $\bar{\sigma} = (\sigma_1,\sigma_2,\ldots,\sigma_{k-1})$.
If you now work out the power $k$ in the left hand side with the multinomial formula, this gives you
$$\sum_{\bar{\sigma}\in S} (-1)^{\sum_{l=1}^{k-1}\sigma_l}\sum_{n_1,\ldots,n_k}\binom{k}{n_1,\ldots,n_k}x_1^{n_1}((-1)^{\sigma_1}x_2)^{n_2}\ldots ((-1)^{\sigma_{k-1}}x_k)^{n_k}$$
Note I didn't write down the condition $n_1+n_2+\ldots+n_k=k$ in the formula to not overload it, but it certainly holds.
The trick is to now change the order of the sums and rearrange all the factors to regroup the $x$'s on one side and the $\sigma$'s on the other. This results in
$$\sum_{n_1,\ldots,n_k}\binom{k}{n_1,\ldots,n_k}x_1^{n_1}x_2^{n_2}\ldots x_k^{n_k}\sum_{\bar{\sigma}\in S} (-1)^{(n_2+1)\sigma_1+(n_3+1)\sigma_2+\ldots+(n_k+1)\sigma_{k-1}}$$
Now, we can focus our attention on that last factor. When all $n_i=1$, you see that the power is even, which means that we always have $1$. Since the cardinality of $S$ is $2^{k-1}$ and the multinomial factor is
$$\binom{k}{1,1,\ldots,1}=k!$$
we get exactly
$$k! 2^{k-1} x_1 x_2 \ldots x_k $$
We now have to understand why all other cases give $0$. But, when at least one of the $n_i\neq 1$, it is inevitable that at least one $n_j=0$, otherwise the sum of them exceeds $k$. But if one of the $n_j$ is $0$, then in the power there is an odd factor $n_j+1$. By the fact we sum over all elements of $S$, we will have as many $+1$ as $-1$ and thus the sum must be $0$.
