<s>[INST] Proof of $d^k + (a-d)^k = a[d^{k-1}-d^{k-2}(a-d)+\dots+(a-d)^{k-1}]$ It is stated to me as part of a larger problem that for odd $a > 2$, positive integer $n$, $k=a^n$, and positive integer $d$, we have
\begin{equation}
d^k + (a-d)^k = a[d^{k-1}-d^{k-2}(a-d)+\dots+(a-d)^{k-1}].
\end{equation}
I have tried expanding but I'm still unsure of how the coefficients on the RHS is equivalent to the binomial expansion on the LHS.
 [/INST] This is just a different form of the factorization for positive odd $n$ $$x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \cdots + y^{n-1}) = (x+y) \sum_{k=0}^{n-1} (-1)^k x^{n-k-1}y^{k}.$$  This is because
$$\begin{align*}
(x+y) \sum_{k=0}^{n-1} (-1)^k x^{n-k-1} y^k 
&= \sum_{k=0}^{n-1} (-1)^k x^{n-k} y^k + \sum_{k=0}^{n-1} (-1)^k x^{n-k-1}y^{k+1} \\
&= x^n + \sum_{k=1}^{n-1} (-1)^k x^{n-k} y^k + \sum_{k=1}^n (-1)^{k-1} x^{n-k} y^k \\
&= x^n + \sum_{k=1}^{n-1} \left( (-1)^k + (-1)^{k-1} \right) x^{n-k} y^k + y^n \\
&= x^n + y^n + \sum_{k=1}^{n-1} 0 \\
&= x^n + y^n.
\end{align*}$$
Your identity follows by setting $n = k$, $x = d$, $y = a-d$.
</s>