We have the arithmetic function $$f(n)=\sum_{d\mid n}\mu (d)\cdot d$$ I want to show that $f\left (p_1^{e_1}\cdots p_k^{e_k}\right )=(-1)^k\cdot (p_1-1)\cdots (p_k-1)$.
We have that $d$ is of the form $p_1^{a_1}\cdots p_k^{a_k}$ with $0\leq a_i\leq e_i$ right?
So $$f\left (p_1^{e_1}\cdots p_k^{e_k}\right )=\sum_{d\mid n}\mu (d)\cdot d=\sum_{0\leq a_i\leq e_i, \forall i}\mu (p_1^{a_1}\cdots p_k^{a_k})\cdot p_1^{a_1}\cdots p_k^{a_k}$$ or not?
How could we continue?
On
A little more general:
Theorem: Suppose $f$ is a multiplicative function. Then $$\sum_{d|n} \mu(d) f(d) = \prod_{p|n} (1-f(p)).$$
Proof: Let $$g(n) = \sum_{d|n} \mu(d) f(d).$$ Then $g$ is multiplicative (the product $\mu f$ is obviously multiplicative so the Direchlet convolution $\mu f * u$ where $u(n) = 1$ for every $n$ is multiplicative). This implies that $g$ is completely determined by its behavior on powers of primes.
That is, $$g(p^k) = \sum_{d|p^k} \mu(d) f(d) = \mu(1) f(1) + \mu(p) f(p) = 1-f(p).$$ Therefore, $$g(n) = \prod_{p|n}g(p^k) = \prod(1-f(p)).$$
Note: Try to deal with the prime powers individually rather than dealing with all of them at once. That is, try handling the case where $n = p^k$ first and then apply the multiplicative property to attain the general case.
Yes, you are correct so far. Now by the multiplicative property of the Mobius function and its definition, it follows that $$\begin{align*}\sum_{0\leq a_i\leq e_i, \forall i}\mu (p_1^{a_1}\cdots p_k^{a_k})\cdot p_1^{a_1}\cdots p_k^{a_k}&= \sum_{0\leq a_i\leq e_i, \forall i}\mu (p_1^{a_1})\cdots \mu(p_k^{a_k})\cdot p_1^{a_1}\cdots p_k^{a_k}\\&= \sum_{0\leq a_i\leq 1, \forall i}(-1)^{a_1}\cdots (-1)^{a_k}\cdot p_1^{a_1}\cdots p_k^{a_k}\\&= \sum_{0\leq a_1\leq 1}(-p_1)^{a_1}\dots\sum_{0\leq a_k\leq 1}\cdots (-p_k)^{a_k}\\&=(1-p_1)\cdots(1-p_k)\\ &=(-1)^k\cdot (p_1-1)\cdots (p_k-1).\end{align*}$$