I'm trying to understand a passage in the proof that e is an irrationale number.
We assume that $e$ is a rational number, $e= \frac{p}{q}$ in which $p$ and $q$ are two coprime numbers.
$$e= \sum_{k=0}^\infty \frac{1}{k!} \implies e-\sum_{k=0}^\infty \frac{1}{k!}=\frac{p}{q}-\sum_{k=0}^\infty \frac{1}{k!}=0 \\\implies \frac{p}{q}-\sum_{k=0}^q \frac{1}{k!}>0 \\\implies 0< \frac{p}{q}-\sum_{k=0}^q \frac{1}{k!}\\ \implies\sum_{k=0}^ \infty \frac{1}{k!} -\sum_{k=0}^q \frac{1}{k!} =\sum_{k=q+1}^\infty \frac{1}{k!} $$
Until this passage it is all clear. Then I don't understand why: $$\sum_{k=q+1}^\infty \frac{1}{k!} <\frac{1}{q!}\cdot \frac{1}{q}$$
Then the proof goes on and it is quite simple to understand.
$$\sum_{k=q+1}^\infty \frac{1}{k!}= \frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots\\ =\frac{1}{q!}\cdot \left( \frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\cdots \right) \\ =\frac{1}{q!}\cdot\frac{1}{q}\cdot \left( \frac{q}{q+1}+\frac{q}{(q+1)(q+2)}+\cdots \right)$$
Why does $\left( \frac{q}{q+1}+\frac{q}{(q+1)(q+2)}+\cdots \right)<1$?
\begin{align}\sum_{k=q+1}^\infty\frac1{k!}&=\frac1{q!}\left(\frac1{q+1}+\frac1{(q+1)(q+2)}+\cdots\right)\\&<\frac1{q!}\left(\frac1{q+1}+\frac1{(q+1)^2}+\cdots\right)\\&=\frac1{q!}\times\frac{\frac1{q+1}}{1-\frac1{q+1}}\\&=\frac1{q!}\times\frac1q.\end{align}