Add the binary numbers 1110 base 2 and 111 base 2.

$0 + 1 = 1$. Write down 1.

$$\begin{array}{c} & 1 & 1 & 1 & 0_2 \\ + & 0 & 1 & 1 & 1_2 \\ \hline & & & & 1 \\ \end{array}$$

$1 + 1 = 10_2$ (which is 0 with a carry of 1). Write down 0, carry 1.

$$\begin{array}{c} & \text{carry } 1 \\ & 1 & 1 & 1 & 0_2 \\ + & 0 & 1 & 1 & 1_2 \\ \hline & & & 0 & 1 \\ \end{array}$$

$1 + 1 + 1 = 11_2$ (which is 1 with a carry of 1). Write down 1, carry 1.

$$\begin{array}{c} \text{carry } 1 & \text{carry } 1 \\ & 1 & 1 & 1 & 0_2 \\ + & 0 & 1 & 1 & 1_2 \\ \hline & & 1 & 0 & 1 \\ \end{array}$$

$1 + 0 + 1 = 10_2$ (which is 0 with a carry of 1). Write down 0, carry 1.

$$\begin{array}{c} \text{carry } 1 & \text{carry } 1 & \text{carry } 1 \\ & 1 & 1 & 1 & 0_2 \\ + & 0 & 1 & 1 & 1_2 \\ \hline & 0 & 1 & 0 & 1 \\ \end{array}$$

$$\begin{array}{c} \text{carry } 1 & \text{carry } 1 & \text{carry } 1 \\ & 1 & 1 & 1 & 0_2 \\ + & 0 & 1 & 1 & 1_2 \\ \hline 1 & 0 & 1 & 0 & 1_2 \\ \end{array}$$

The sum is $10101_2$.

Comparing this result with the given options:

(a) $1111_2$

(b) $10100_2$

(c) $10101_2$

(d) $110_2$

(e) $10011_2$

The correct option is (c).

The final answer is $\boxed{\text{(c) } 10101_2}$.