Induced connection on exterior powers is a degree zero derivation

Realize ${\bigwedge }^{k}E$ as the alternating tensors in $E^{\otimes k}$, or alternating multilinear forms on $(E^{\ast }{)}^k$. Locally the signed sums ${\sum }_{\sigma \in S_k}\mathrm{sgn}(\sigma )e_{i_{\sigma (1)}}\otimes \cdots \otimes e_{i_{\sigma (k)}}$ for $i_1<\cdots <i_k$ form a basis: alternation makes repeated-index coefficients zero and determines all distinct-index coefficients from the increasing ones. Frame changes preserve alternation, with smooth polynomial coefficients and smooth inverses. Hence these local frames define a smooth subbundle. The normalized alternating projection is ${\mathrm{Alt}}_k=(1\mathrm{/}k!){\sum }_{\sigma }\mathrm{sgn}(\sigma )\sigma$; alternation gives ${\mathrm{Alt}}_k^2={\mathrm{Alt}}_k$, and its image is exactly this subbundle.

Since ${\mathrm{\nabla }}_X$ commutes with every permutation and real constant, it commutes with ${\mathrm{Alt}}_k$. It therefore preserves the image and its restriction obeys the connection laws. Write the usual wedge of alternating tensors as $S\wedge T=((p+q)!\mathrm{/}(p!q!)){\mathrm{Alt}}_{p+q}(S\otimes T)$. Applying the product rule and commuting the alternating projection with the derivative gives the claimed sum with two positive signs.

Degree zero uses $0!=1$ and ordinary scalar multiplication, so the formula becomes the ordinary scalar Leibniz rule. For $k>\mathrm{rank}E$ there is no increasing index tuple and the bundle is zero; a rank-zero bundle consequently has only its degree-zero scalar part. Degree one returns $E$. Empty bases present no sections to test. The finite permutation sums and local frames require no AC.