Your hypothesis is in a sense stronger than just assuming ZFC outright. 

Namely, if we have $\lambda$-DC for some inaccessible cardinal $\lambda$, and ZF in the background, then in particular, we will have the full axiom of choice inside the universe $V_\lambda$, consisting of all sets of rank less than $\lambda$. So $V_\lambda$ will be a model of ZFC. 

Therefore, by simply jumping inside this universe $V_\lambda$, we recover full ZFC for all the purposes of "ordinary mathematics." 

But let me note that it is somewhat subtle to define what one means by inaccessible cardinal in the absence of the axiom of choice, since the usual definition would be that $\lambda$ is an uncountable regular strong limit, but being a strong limit should mean that if $\kappa<\lambda$ then $P(\kappa)<\lambda$ as well, and so in particular, $P(\kappa)$ is well-orderable. It follows that if $\lambda$ is inaccessible, then the axiom of choice holds in $V_\lambda$ just as a consequence of inaccessibility. 

In other words, we get ZFC in $V_\lambda$ even without your $\lambda$-DC assumption. In this sense, the power of your hypothesis is having an inaccessible cardinal, rather than in having $\lambda$-DC.