Prove that
$$\sum_{q=0}^v \binom{v}{q}\frac{q!}{v^{q+1}} = \sum_{q=0}^{v-1} \binom{v-1}{q} \frac{(q+2)!}{v^{q+2}}$$
Thanks.
Below are what I have tried:
Approach 1:
$$\sum_{q=0}^{v-1} \binom{v-1}{q} \frac{(q+2)!}{v^{q+2}}-\sum_{q=0}^v \binom{v}{q}\frac{q!}{v^{q+1}} \\ =\sum_{q=1}^{v} \binom{v-1}{q-1} \frac{(q+1)!}{v^{q+1}}-\sum_{q=1}^v \binom{v}{q}\frac{q!}{v^{q+1}}-\frac{1}{v}\\ =\sum_{q=1}^{v} \binom{v-1}{q-1} \frac{(q-1)!}{v^{q+1}}(q^2+q-v)-\frac{1}{v} \\ =(v-1)!\sum_{q=1}^{v} \frac{q^2+q-v}{v^{q+1}(v-q)!}-\frac{1}{v} \\ $$ It is shown by Mathematica that
$$ \sum_{q=1}^{v} \frac{q^2+q-v}{v^{q+1}(v-q)!}=\frac{1}{v!}, (Eq.1)\\ $$
then the equation follows. However, I failed to mathematically prove Eq.1.
Approach 2:
Trying to predict the summation result, e.g., $A(v)$. Prove that both the summations on the left and right are equal to $A(v)$ using mathematical induction. However, I cannot figure out $A(v)$.