You Forgot Five!

To find e:
Add all equations up to get 4a+4b+4c+4d+4e=16
Simplify: a+b+c+d+e=4
Subtract:
a+b+c+d+e=4

a+b+c+d=1

e=3

To find d with e:
first equation-fifth equation:
a+b+c+d=1

a+b+c=3

d=-2

To find a with d and e:
fifth equation-second equation:
a+b+c=3

b+c=1

a=2

To find b with a, d, and e:
Use fourth equation
-2+3+2+b=4
b=1

To find c with a, b, d, and e
Use fifth equation
3+2+1=6
c=0

Adding all $5$ sums gives $4*(a+b+c+d+e) = 16 \Rightarrow a+b+c+d+e=4 \quad (1)$ From this we can subtract the sums from $(1)$ (for example we would say $(a+b+c+d+e) - (a+b+c+d) = 4-1 \Rightarrow e = 3$ ) to obtain the follwing results $a=2, \quad b=1, \quad c=0, \quad d=-2, \quad e=3$ Therefore $a^e + b^d + c = 2^3 + 1^{-2} + 0 = 8 + 1 = \fbox{9}$

If we sum all the equations then we get $4(a+b+c+d+e) = 16$ . Therefore $a+b+c+d+e = 4$ . Subbing in the initial equations gives us 5 new equations: $a+2 = 4$ $b+3 = 4$ $c+4 = 4$ $d+6 = 4$ $e+1 = 4$ Solving these gives us the solution $(2,1,0,-2,3)$ so $a^e+b^d+c = 2^3+1^{-2}+0= 8+1 = \fbox{9}$