Vtyjkyuk

Let $m,n$ be positive integers and $N=1,ldots, n$. Try to prove the two following identites:

1. For $m<n$, we have
   $$sum_A subseteq N (-1)^ Aright left(sum_j in A x_jright)^m = 0.$$

E.g., for $n=3, m=1$, we have
$$(x_1+x_2+x_3) = (x_1+x_2) + (x_1+x_3) + (x_2+x_3) - x_1 - x_2 - x_3,$$
and for $n=3, m=2$, we have again
$$(x_1+x_2+x_3)^2 = (x_1+x_2)^2 + (x_1+x_3)^2 + (x_2+x_3)^2 - x_1^2 - x_2^2 - x_3^2.$$

2. For $m=n$, we have
   $$sum_A subseteq N (-1)^ Aright left(sum_j in A x_jright)^n = (-1)^n n! prod_i=1^nx_i.$$

E.g., for $m=n=2$, we have
$$(x_1+x_2)^2 = x_1^2 + x_2^2 + 2!x_1 x_2,$$
and for $m=n=3$, we have
$$(x_1+x_2+x_3)^3 = (x_1+x_2)^3 + (x_1+x_3)^3 + (x_2+x_3)^3 - x_1^3 - x_2^3 - x_3^3 + 3!x_1x_2x_3.$$

I think the both identities could be proved by mathematical induction, but I still have no idea how to prove them. Hoping to see some smart ideas. Thanks a lot!

First note that
$$sum_A subseteq N (-1)^ Aright left(sum_j in A x_jright)^mtag 1$$
is an homogeneus polynomial of degree $m$ in the variables $x_1,ldots,x_n$.
Let consider a generic monomial $prod_i=1^nx_i^e_i$ with $e_iinBbb N$ and $sum_i=1^ne_i=m$.
Its coefficient in $(1)$ is given by
$$sum_Ssubseteq Asubseteq N(-1)^fracm!prod_i=1^ne_i!$$
where $S=i:e_i>0$.
Moreover
beginalign
sum_Ssubseteq Asubseteq N(-1)^
&=(-1)^Ssum_Bsubseteq Nsetminus S(-1)^\
&=(-1)^Ssum_k=0^binomk(-1)^k\
&=
begincases
0&Sneq N\
(-1)^S&S=N
endcases
endalign

If $m<n$ or ($m=n$ and $e_i>1$ for some $i$), then $Sneq N$, hence $(1)$ is proved to be $0$ in this case.

If $m=n$ and $e_i=1$ for all $i$, then $S=N$, hence the coefficient is $(-1)^nn!$ and this proves the second identity.