$$\begin{align}a+b+c+d&=1\\ a^2+b^2+c^2+d^2&=2\\ a^3+b^3+c^3+d^3&=3\\ a^4+b^4+c^4+d^4&=4\\ a^5+b^5+c^5+d^5&- ?\end{align}$$
The usual method I see for solving this kind of problems is bashing using the Newton sums.
Let $P_n=a^n+b^n+c^n+d^n$.
If we take $a,b,c,d$ to be the roots of
$$f(x)=x^4-x^3-\frac{1}{2}x^2-\frac{1}{6}x+\frac{1}{24},$$
then they satisfy all the initial conditions and we also have $P_5=\frac{139}{24}$.
But what this proves is that there exist $a,b,c,d$ such that the initial conditions are satisfied and $P_5=\frac{139}{24}$.
This alone does not prove that this is the only possible value of $P_5$. How to prove that it is, generally?
(It might be possible to find the value using only interesting kinds of factorization identities, which would show that the value is unique, but such strategy would not be as general as the Newton sums)
In addition you simply need to note that there is a unique monic quartic polynomial $(x-a)(x-b)(x-c)(x-d)=0$ which has the four roots $a,b,c,d$
Suppose $a,b,c,d$ are the roots of the polynomial $f(x)=x^4+px^3+qx^2+rx+s=0$ - then $p,q,r,s$ are uniquely defined in terms of $a,b,c,d$ and indeed in terms of the given sums of powers.
Now compute $$0=af(a)+bf(b)+cf(c)+df(d)=a^5+b^5+c^5+d^5+4p+3q+2r+s$$on substituting the known values for $a+b+c+d, a^2+b^2+c^2+d^2$ etc
Thus the sum of fifth powers is uniquely determined by $p,q,r,s$ which are uniquely determined by the given data.