Find the value of $\sum_{n=1}^{100} i^n$.

The sum is $i^1 + i^2 + i^3 + i^4 + \dots + i^{100}$.
We know that the sum of any four consecutive powers of $i$ is $0$ because $i + (-1) + (-i) + 1 = 0$.
The total number of terms is 100, which is perfectly divisible by 4 ($100 = 4 \times 25$).
Therefore, there are exactly 25 blocks of 4 terms, each summing up to 0.

$$Sum = 25 \times 0 = 0$$

Answer: 0