How is it possible to get this equality? What can one do for this ? I find it hard to see how starting from $k=1$ results in a binomial that has got $k-1$.
$\sum_{k=0}^{n-1} \binom{n}{k} a^{n-k} b^{k+1} = \sum_{k=1}^{n} \binom{n}{k-1} a^{n-(k-1)} b^{(k-1)+1}$
They reindexed the summation. Starting with $$\sum_{k=0}^{n-1} \binom{n}{k}a^{n-k}b^{k+1},$$ set $l = k+1 \implies l-1 = k.$ When we index in terms of $l$, we are summing from $1$ to $n$. The lower index of the sum isn't notationally correct but I wrote it in to have it make sense. $$\sum_{k=0}^{n-1} \binom{n}{k}a^{n-k}b^{k+1} = \sum_{l-1=0}^{(l-1=)n-1} \binom{n}{l-1}a^{n-(l-1)}b^{(l-1)+1} = \sum_{l=1}^n \binom{n}{l-1}a^{n+1-l}b^l.$$