I'm almost sure that I haven't understood correctly the Poisson formula
$$\sum_{n=-\infty}^{+\infty}f(t-nT)=\frac{1}{T}\sum_{m=-\infty}^{+\infty}F\left(\frac{2\pi m}{T}\right) e^{\frac {-i2\pi m t}{T}}.$$
If I try to apply it to the shifted delta, I get:
$$\sum_{n=-\infty}^{+\infty}\delta(t-nT)=\frac{1}{T} \sum_{m=-\infty}^{+\infty} e^{\frac {-i2\pi m (nT)}{T}}e^{\frac {-i2\pi m t}{T}}$$
but I have obtained two different indices (m and n) in the same sum.... (right hand of expression)
The correct answer - if I'm not wrong - should be $$\sum_{n=-\infty}^{+\infty}\delta(t-nT)=\frac{1}{T} \sum_{m=-\infty}^{+\infty} e^{\frac {-i2\pi m t}{T}}$$
Thanks for your help