Sure, here is the translated text:

```
Simplify the following expression:

\[
\begin{aligned}
& 1+\frac{a_{1}}{1-a_{1}}+ \frac{a_{2}}{\left(1-a_{1}\right)\left(1-a_{2}\right)}+\frac{a_{3}}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)}+ \\
& \frac{a_{4}-a_{1}}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)\left(1-a_{4}\right)}
\end{aligned}
\]
```
First, add the first two terms of the expression, then add the third term to their sum, the fourth term to the obtained sum, and finally the fifth term. In this way, we successively obtain the following expressions:

$$
\begin{aligned}
& 1+\frac{a_{1}}{1-a_{1}}=\frac{1}{1-a_{1}}, \\
& \frac{1}{1-a_{1}}+\frac{a_{2}}{\left(1-a_{1}\right)\left(1-a_{2}\right)}=\frac{1}{\left(1-a_{1}\right)\left(1-a_{2}\right)} \\
& \frac{1}{\left(1-a_{1}\right)\left(1-a_{2}\right)}+\frac{a_{3}}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)}=\frac{1}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)}, \\
& \frac{1}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)}+\frac{a_{4}-a_{1}}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)\left(1-a_{4}\right)}= \\
& \frac{1-a_{1}}{\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right)\left(1-a_{4}\right)}
\end{aligned}
$$

The last sum can be simplified with the numerator:

$$
\frac{1}{\left(1-a_{2}\right)\left(1-a_{3}\right)\left(1-a_{4}\right)}
$$

$\left(1-a_{1} \neq 0\right.$, otherwise the original expression would not make sense.)