A different approach that also uses a Poisson distribution with parameter $\lambda t$ would be:
$P(t < T_n < t+dt) = \underbrace{ P(N(t) = n-1)}_{\substack{\text{the probability for }\\\text{$(n-1)$ arrivals at time $t$}}} \qquad \qquad \times \qquad \underbrace{\vphantom{P(N(t) = n-1)}\lambda dt}_{\substack{\text{the probability for }\\\text{arrival between time $t$ and $t+dt$}}} $
leading to
$P(t < T_n < t+dt)= \text{Poisson}(n-1,\lambda t) \times \lambda dt = \lambda\frac{(\lambda t)^{n-1} e^{-\lambda t}}{(n-1)!} dt = \frac{\lambda e^{-\lambda t}(\lambda t)^{n-1} }{\Gamma(n)} dt$
The connection with your approach is as following:
You could view the derivative as $\frac{d}{dt} P(N(t)
and
$\underbrace{{\frac{d}{dt} P(N(t)=k)}}_{\text{change of k}} = \underbrace{\lambda P(N(t)=k-1)\vphantom{\frac{d}{dt} P(N(t)=k)}}_{\text{gain from k-1 to k}} -\underbrace{\lambda P(N(t)=k)\vphantom{\frac{d}{dt} P(N(t)=k)}}_{\text{loss from k to k+1}}$
and all those terms cancel (much like the answer of Sasha):
$\frac{d}{dt} P(N(t)
So the the derivative of $P(N(t) < n)$ the rate at which you are surpassing the 'level' $n-1$ is equal to how many there is currently at the 'level' $n-1$ and how fast those will rise to the 'level' $n$. What happens at lower 'levels' does not matter for the change of $P(N(t) < n)$.
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