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My first response was wrong, so let me correct it. I claim now that indeed $$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n\asymp_\mu\frac{1}{nt}. $$ For the lower bound observe that the largest term occurs for $\mu^k\asymp_\mu \frac{1}{nt}$ and it contributes $\asymp_\mu\frac{1}{nt}$ to the sum. For the upper bound observe that the function $x\mapsto \mu^x (1-\mu^x t)^n$ increases when $\mu^x>\frac{1}{nt}$, decreases when $\mu^x<\frac{1}{nt}$, and has a maximum $\asymp\frac{1}{nt}$ when $\mu^x=\frac{1}{nt}$. Therefore, by the familiar technique of comparing an infinite sum to an integral, we obtain $$ \sum_{k=0}^\infty\mu^k(1-\mu^kt)^n \ll \frac{1}{nt}+\int_0^\infty \mu^x (1-\mu^x t)^n\ dx \ \ll_\mu \frac{1}{nt}.$$ Note that the integral can be evaluated explicitly by the change of variable $s:=\mu^x$, say.