Laplace transform of a sampled signal
Solution 1:
$(1-e^{-sT})F(s)$ is the transform of $g(t)=f(t)-f(t-T)$. The sampled signal $$ g^*(t)=\sum [f(kT)-f((k-1)T)]\delta(t-kT) $$ has again the form of a difference $f^*(t)-f^*(t-T)$. This only works because the shift is equal to the sampling period, else the values in the second term would be drawn from a different sample sequence. So in the end you get $$ [(1-e^{-sT})F(s)]^*=(1-e^{-sT})F^*(s). $$