First: You probably meant to write $[1 - (1 - e^{ - \alpha } )/\alpha ]^5$ as the solution to this exercise, right?
Hint 1: The arrival times of the five customers can be simulated as five i.i.d. uniform$[0,1]$ rv's. That is, if you place five i.i.d. uniform rv's on $[0,1]$, the points correspond to arrival times of five customers.
Hint 2: If $Z_1,\ldots,Z_5$ are i.i.d. rv's with distribution function $F$, how can you express the distribution function of $\max \{ Z_1 , \ldots ,Z_5 \}$ in terms of $F$?
Hint 3: You'll have to use the law of total probability, in order to calculate a certain probability. In this context, it may be useful to note that $1-U$ and $U$ are identically distributed, for $U$ a uniform$[0,1]$ random variable.
EDIT (further hints, in response to the OP's request). Suppose that $Z = U + Y$, where $U$ and $Y$ are independent uniform$[0,1]$ and exponential$(\alpha)$ random variables, respectively. First note that
$$
{\rm P}(Z \le 1) = {\rm P}(U + Y \le 1) = {\rm P}(Y \le 1 - U) = {\rm P}(Y \le U),
$$
where the last equality (which is actually not essential) follows from the fact that $U$ and $1-U$ are identically distributed. Now, since $U$ has constant density $f(u)=1$ for $u \in [0,1]$, the law of total probability gives
$$
{\rm P}(Y \le U) = \int_0^1 {{\rm P}(Y \le U|U = u)1\,{\rm d}u} = \int_0^1 {{\rm P}(Y \le u)\,{\rm d}u} = \int_0^1 {(1 - e^{ - \alpha u} )\,{\rm d}u} .
$$
Calculate the integral on the right-hand side, and recall Hint 2.
