I saw a proof that makes this step without any additional explanation:
$$\sum_{t=0}^{k-1} \binom{k}{k-1-t}(n-1)^{k-1-t} = \sum_{t=0}^{k-1} \binom{k}{t}(n-1)^{t} $$
I know that $ \binom{k}{k-1-t} = \binom{k}{t+1} $ by the symmetry property. I also try just expanding and got $$\sum_{t=0}^{k-1} \binom{k}{t}(n-1)^{t} \frac{k-t}{1+t} (n-1)^{k-1-2t} $$ but I can't find a reason for $\frac{k-t}{1+t} (n-1)^{k-1-2t} $ to be ignorable here.
Any hints?
This follows from $$\sum_{t=0}^{k-1}a_{k-1-t}=\sum_{t=0}^{k-1}a_t,$$ which holds for any sequence $\{a_t\}_{0\leq t\leq k-1}$.
Indeed, we have $$\sum_{t=0}^{k-1}a_{k-1-t}=a_{k-1}+a_{k-2}+\cdots+a_1+a_0$$ and $$\sum_{t=0}^{k-1}a_t=a_0+a_1+\cdots+a_{k-2}+a_{k-1}.$$ So they are both the sum of the sequence $\{a_t\}_{0\leq t\leq k-1}$ and thus equal.