Proof by induction: the $n$-th derivative $f^{(n)}$ of function $f(x)=xe^{x-1}$ is same as ... 
Prove by induction that for all $n \in \mathbb{N}, n \geq 1$ the
  $n$-th derivation $f^{(n)}$ of the function $f: \mathbb{R} \rightarrow
\mathbb{R}, f(x) = x \cdot e^{x-1}$ is defined / same as  (translated
  word-by-word: "given by") $f^{(n)}(x) = (x+n)e^{x-1}$.

For $n=1:$ (This means we take $f$ aka the given function and create its first derivative:)
$$(x\cdot e^{x-1})' = e^{x-1}+x\cdot e^{x-1} = (1+x) \cdot e^{x-1}$$
Now we take the other thing and put $n=1:$
$$f^{(1)}(x) = (x+1)e^{x-1}$$
We see it's same, so far so good.

Now we prove for $n+1$ but I have problem with notation because this task is very very stupid made imo.
$$f^{(n+1)}(x) = (f^{(n)})'(x)$$
$$f^{(n+1)}(x) = ((x+n)e^{x-1})'$$
No idea if I used correct notation this task gives me massive headaches it's from old exam and makes me very mad because the notation is very confusing here. I don't know if I did it correct till here...
Any ideas how to make the correct notation and end the proof?

You could always bypass having to use the $'$ symbol and the $^{(n)}$ together in the same breath.
To prove the inductive implication 
$f^{(n)}(x) = (x+n)e^{x-1} \Longrightarrow f^{(n+1)}(x)=(x+(n+1))e^{x-1},$ 
why not assume the inductive hypothesis true, and then set $g = f^{(n)}$, so that $g(x) = (x+n)e^{x-1}$. The desired conclusion can now be stated as
$f^{(n+1)}(x) = g'(x)=(x+(n+1))e^{x-1}$,
where the first of these equalities is due simply by definition of $g$, and the second equality is what actually needs to be proven.
BTW your original use of notation is correct - just not neat, which is more due to the limitations of prime notation than your use of it.
