$ P(N_t-N_s=0 \mid N_t>0) = \frac{P(N_t-N_s=0\ \&\ N_t>0)}{P(N_t>0)} = \frac{P(N_t-N_s=0\ \&\ N_s>0)}{P(N_t>0)}. $ First figure out what the above is true. Think about what it means.
Then exploit the fact that the two events with "$\&$" between them are independent. Make sure you understand why they're independent.
Later note in response to comments below: What was done above was for the purpose of writing the expression in terms of events that are independent. $ \Pr(N_t-N_s=0) = e^{-(t-s)\lambda} $ $ \Pr(N_s>0) = 1 - e^{-s\lambda} $ $ \Pr(N_t>0) = 1 - e^{-t\lambda} $ So we have $ \frac{e^{-(t-s)\lambda}(1 - e^{-s\lambda})}{1 - e^{-t\lambda}}. $ Multiplying out the numerator, we get: $ \frac{e^{-(t-s)\lambda} - e^{-t\lambda}}{1 - e^{-t\lambda}}. $ Then multiply both the numerator and the denominator by $e^{t\lambda}$, and we get: $ \frac{e^{s\lambda}-1}{e^{t\lambda}-1}. $