Let $t_0,t_1,\ldots,t_m$ be variables, let $k$ be a non-negative integer. Denote also $[m]=\{1,2,\ldots,m\}$.
Fact. The following identity holds: $$ \sum_{I\subset [m]}(-1)^{m-|I|}\left(t_0+\sum_{i\in I}t_i\right)^k=\sum_{\substack{0\le i_1,\ldots,i_k\le m\\ [m]\subset\{i_1,\ldots,i_k\}}}t_{i_1}\ldots t_{i_k}. $$ Example. For $m=2$ and $k=3$ we have $$ (t_0+t_1+t_2)^3-(t_0+t_1)^3-(t_0+t_2)^3+t_0^3=6t_0t_1t_2. $$
Basically the identity says that by taking this alternating sum one can extract from multinomial $(t_0+t_1+\ldots+t_m)^k$ all terms that contain at least one $t_1$, at least one $t_2$ and so on. In other words, the right hand side can be rewritten as $$ \sum_{\substack{d_0\ge 0,\,d_1\ldots,d_m\geq 1\\ d_0+d_1+\ldots+d_m=k}}t_{0}^{d_0}t_{1}^{d_1}\ldots t_{m}^{d_{m}}, $$ which is part of the multinomial.
The identity can be proved in various ways (e.g. induction, via difference operators), but my question is whether this inclusion-exclusion-like identity can be actually deduced from the inclusion-exclusion formula.
Let me now show some ideas in this direction. Let $A=\{t_0,t_1,\ldots,t_m\}$ and $A_{i}=A\setminus\{t_i\}$ for $1\le i\le m$. For each subset $S\subset A$ define $$ f(S)=\sum_{x_1,\ldots,x_k\in S}x_1x_2\ldots x_k=\left(\sum_{x\in S}x\right)^k. $$ Then, the left hand side of the identity can be rewritten as $$ f(A)-\sum_{i}f(A\cap A_i)+\sum_{i<j}f(A\cap A_i\cap A_j)-\ldots=\sum_{I\subset[m]}(-1)^{|I|}f\left(A\cap\bigcap_{i\in I}A_i\right) $$ Note the similar looking identity (this is basically the inclusion-exclusion formula) $$ |A|-\sum_{i}|A\cap A_i|+\sum_{i<j}|A\cap A_i\cap A_j|-\ldots=\sum_{I\subset[m]}(-1)^{|I|}\left|A\cap\bigcap_{i\in I}A_i\right|=|A\setminus (A_1\cup A_2\cup\ldots\cup A_m)|. $$ Now let us proceed to the proof of the identity (one can notice that the inclusion-exclusion formula above can be shown in the same way). The sketch is as follows: take any monomial $x_1\ldots x_k$, where $\{x_1,\ldots,x_k\}\subset A$ and denote $X=\{x_1,\ldots,x_k\}\setminus\{t_0\}\subset\{t_1,\ldots,t_m\}$, then in the summation above this monomial is "counted" with sign $(-1)^{|S|}$ for all subsets $S\subset\{t_1,\ldots,t_m\}$ such that $X\cap S=\varnothing$. Thus, the total coefficient of $x_1\ldots x_k$ is one if $X=\{t_1,\ldots,t_m\}$ and zero otherwise which proves the identity.
However, I think that it is possible to derive the identity directly from the inclusion-exclusion formula itself (instead of repeating the same argument). Is it really possible?
One possible approach is to pass to the indicator functions of subsets and "integrate" some function (i.e. consider expressions of the form $\int_{A\cap A_i}f$), but I do not see how apply it here.
An help would be appreciated.