Relation of complete homogeneous symmetric polynomials and the elementary symmetric polynomials I was reading about the symmetric polynomials and saw the following relation: 

$$\sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},\ldots ,X_{n})h_{{m-i}}(X_{1},\ldots ,X_{n})=0\text{ for } m>0$$

The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials
SECOND EDITION I. G. MACDONALD, page: 21.
(Question) I was wondering if this relation still holds for any length $k$, i.e. 

$$\sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},\ldots ,X_{n})h_{{m-i}}(X_{1},\ldots ,X_{k})=0\text{ for }k \in \{1,...,n\}$$

P.S. I tried cases and it holds for any case.

We have
$$h_p(X_1,\ldots,X_n)=\sum h_r(X_1,\ldots,X_k)h_{p-r}(X_{k+1},\ldots,X_n)$$
$$e_p(X_1,\ldots,X_n)=\sum e_r(X_1,\ldots,X_k)e_{p-r}(X_{k+1},\ldots,X_n)$$
$$\sum_{i=0}^m\sum_{p=0}^i\sum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,\ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,\ldots,X_k)e_p(X_{k+1},\ldots,X_n)h_q(X_{k+1},\ldots,X_n)$$
If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want. 
