While trying to understand a paper of Cayley, he left something unexplained, I managed to show that it is equivalent to the following formula, which I got stuck at:

$k \cdot (f^k)^{(k-1)} = \sum_{j=0}^{k-1} {{k} \choose {j}} (f^{j})^{(j)}(f^{k-j})^{(k-j-1)} $

Can somebody help?

While trying to understand a paper of Cayley, he left something unexplained, I managed to show that it is equivalent to the following formula, which I got stuck at:

$$k \cdot (f^k)^{(k-1)} = \sum_{j=0}^{k-1} {{k} \choose {j}} (f^{j})^{(j)}(f^{k-j})^{(k-j-1)}. $$

Can somebody help?

While trying to understand a paper of caley, he left something unexplained, I managed to show that it is equivalent to the following formula, which I got stuck at:

$k \cdot (f^k)^{(k-1)} = \sum_{j=0}^{k-1} {{k} \choose {j}} (f^{j})^{(j)}(f^{k-j})^{(k-j-1)} $

Can somebody help?

While trying to understand a paper of Cayley, he left something unexplained, I managed to show that it is equivalent to the following formula, which I got stuck at:

$k \cdot (f^k)^{(k-1)} = \sum_{j=0}^{k-1} {{k} \choose {j}} (f^{j})^{(j)}(f^{k-j})^{(k-j-1)} $

Can somebody help?