Remainder of function

We can find the last digit of each of $1^n, 2^n, 3^n, 4^n, f(n)$ as follows:

$\begin{array}{c|c|c|c|c|c} n & 1^n & 2^n & 3^n & 4^n & f(n) \\ \hline \\ 1 & \mathbf{\color{#3D99F6}1} & \mathbf{\color{#3D99F6}2} & \mathbf{\color{#3D99F6}3} & \mathbf{\color{#3D99F6}4} & 1+2+3+4 = 1\mathbf{\color{#3D99F6}0} \\ 2 & 1\cdot 1 = \mathbf{\color{#3D99F6}1} & 2\cdot 2 = \mathbf{\color{#3D99F6}4} & 3\cdot 3 = \mathbf{\color{#3D99F6}9} & 4 \cdot 4 = 1\mathbf{\color{#3D99F6}6} & 1+4+9+6 = 2\mathbf{\color{#3D99F6}0} \\ 3 & 1\cdot 1 = \mathbf{\color{#3D99F6}1} & 2\cdot 4 = \mathbf{\color{#3D99F6}8} & 3\cdot 9 = 2\mathbf{\color{#3D99F6}7} & 4\cdot 6 = 2\mathbf{\color{#3D99F6}4} & 1+8+7+4 = 2\mathbf{\color{#3D99F6}0} \\ 4 & 1\cdot 1 = \mathbf{\color{#3D99F6}1} & 2\cdot 8 = 1\mathbf{\color{#3D99F6}6} & 3\cdot 7 = 2\mathbf{\color{#3D99F6}1} & 4\cdot 4 = 1\mathbf{\color{#3D99F6}6} & 1+6+1+6 = 1\mathbf{\color{#3D99F6}4} \\ 5 & 1\cdot 1 = \mathbf{\color{#3D99F6}1} & 2\cdot 6 = 1\mathbf{\color{#3D99F6}2} & 3\cdot 1 = \mathbf{\color{#3D99F6}3} & 4\cdot 6 = 2\mathbf{\color{#3D99F6}4} & 1+2+3+4 = 1\mathbf{\color{#3D99F6}0} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{array}$

Here, we note that the last digits in the row for $n=5$ are identical to those in the row for $n=1,$ so if we continued, the rows for $n=6$ and $n=2$ would be identical, etc. That is, this shows that the chart above will repeat every four rows. This tells us that

$\text{Last digit of } f(n) = \begin{cases}4 & \text{ if } n \text{ is a multiple of } 4 \\ 0 & \text{ otherwise} \end{cases}$

from which we can conclude the last digit of $f(1) + f(2) + f(3) + f(4) + \cdots + f(2018)$ is the same as the last digit of $4 \cdot \#\{\text{positive multiples of } 4 \text{ which are } \leq 2018\} = 4\left\lfloor\frac{2018}{4}\right\rfloor = 4\cdot 504 = 201\mathbf{\color{#3D99F6}6}$ giving an answer of $\boxed{6}$