I'll quickly setup the question. This is from a microeconomics text covering the contractual relationship between the owner of a store and the manager, who will operate the store.
$$
\pi_g = e + \epsilon
$$
$ \pi_g $ - Gross Profit
$ e $ - The manager's effort
$ \epsilon $ - Random Error
Taking the abstraction one step further, $$ \pi_{n}=\pi_{g}-s, $$
where, $ \pi_n $ represents Net Profit, s is the manager's salary.
It's assumed error will be mean zero and variance $ \sigma^2 $ such that, the owner will maximize the expected value of his net profit:
$$ E\left(\pi_{n}\right)=E(e+\varepsilon-s)=e-E(s) $$
Question: Mathematically speaking, how does the function simplify to e instead of expectation of e?
This is the nature of expectation. $e$ is a constant. And the expectation of an constant is just the constant itself.
More generally. $\mathbb E(X+c)=\mathbb E(X)+\mathbb E(c)$, where X is a random variable with expectation $\mu$. Here linearity of expectation has been applied. Then we use the the fact, that the expectation of an constant is the constant itself.
$$\mathbb E(X)+\mathbb E(c)=E(X)+0=\mu+0=\mu$$
In your case we have
$$E\left(\pi_{n}\right)=\mathbb E(e+\varepsilon-s)=\mathbb E(e)+\mathbb E(\varepsilon)-\mathbb E(s)=e+0-\mathbb E(s)=e-\mathbb E(s)$$
You probably interested in the variance. The constant has no impact on the variance of the sum. $$Var(e+\varepsilon-s)=Var(\varepsilon-s)$$
It seems that $\varepsilon$ and $s$ are independent. That means $Cov(\varepsilon,s)=0$. Therefore $$Var(\varepsilon-s)=Var(\varepsilon)+Var(s)=\sigma^2+Var(s)$$