Here is a picture to add to Robert's good answer demonstrating the "re-use" of operations, in this case for an 8 point DFT. The "Twiddle Factors" are represented in the diagram using the notation $W_N^{nk}$ which is equal to $e^{j2\pi \frac{nk}{N}}$

Note the path shown and the equation underneath shows the result for the frequency bin X(1), as given by Robert's equation.

Dashed lines are no different than solid lines just to make clear where the summation joins are.

Here is a picture to add to Robert's good answer demonstrating the "re-use" of operations, in this case for an 8 point DFT. The "Twiddle Factors" are represented in the diagram using the notation $W_N^{nk}$ which is equal to $e^{j2\pi \frac{nk}{N}}$

Note the path shown and the equation underneath shows the result for the frequency bin X(1), as given by Robert's equation copied below:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{j 2 \pi \frac{nk}{N}} $$

Dashed lines are no different than solid lines just to make clear where the summation joins are.

Here is a picture to add to Robert's good answer demonstrating the "re-use" of operations, in this case for an 8 point DFT. The "Twiddle Factors" are represented in the diagram using the notation $W_N^{nk}$ which is equal to $e^{j2\pi \frac{nk}{N}}$

Note the path shown and the equation underneath shows the result for the frequency bin X(1), as given by Robert's equation.

Here is a picture to add to Robert's good answer demonstrating the "re-use" of operations, in this case for an 8 point DFT. The "Twiddle Factors" are represented in the diagram using the notation $W_N^{nk}$ which is equal to $e^{j2\pi \frac{nk}{N}}$

Note the path shown and the equation underneath shows the result for the frequency bin X(1), as given by Robert's equation.

Dashed lines are no different than solid lines just to make clear where the summation joins are.