## show that the differential of a differential form is linear

having two differential forms
$\displaystyle \omega:=\displaystyle\sum_{i_{1}<...<i_{k}}\omega_ {i_{1},...,i_{k}}dx^{i_{1}}\wedge...\wedge dx^{i_{k}}$and its differential:$\displaystyle d\omega=\displaystyle\sum_{i_{1}<...<i_{k}}d\omega _{i_{1},...,i_{k}}\wedge dx^{i_{1}}\wedge...\wedge dx^{i_{k}}$

$\displaystyle \eta:=\displaystyle\sum_{j_{1}<...<j_{l}}\eta_{j_{ 1},...,j_{l}}dx^{j_{1}}\wedge...\wedge dx^{j_{l}}$ then $\displaystyle d\eta=\displaystyle\sum_{j_{1}<...<j_{l}}d\eta_{j_ {1},...,i_{l}}\wedge dx^{j_{1}}\wedge...\wedge dx^{j_{l}}$

can someone show me how to prove that the differential is linear? The proof is said to be trivial,...but I have problems with showing it
$\displaystyle d(\omega+\eta)=d\omega+d\eta$