I've recently started studying recursive definitions of sequences and have encountered a recurrence formula which seems to include a factorial. Until now I have not seen any like this.
$$t_n = 5 \,\frac{(n + 1)!}{4^n} ,\quad n ≥ 1.$$
If I was to take a look at this segment (n + 1)! and replace n with position 1 would the factorial simply be 2! which I can then input back into the recurrence formula like :
$$t_n = 5\, \frac{2!}{4^n} ,\quad n ≥ 1.$$
Apologies for any incorrect terminology.
$$T_n=5 \frac{(n+1)!}{4^n}~~~~(1)$$ will follow the recurrence relation given by $$T_{n}=\frac{1}{4}(n+1)T_{n-1}~~~~(2)$$
EDIT for OP:
Now in order to get (1) from (2) one can do multiplicative telescoping or divide (2) by $(n+1)!$ to get $$\frac{T_n}{(n+1)!}=\frac{1}{4}\frac{(n+1)T_{n-1}}{(n+1)!}=\frac{1}{4}\frac{T_{n-1}}{n}~~~(3)$$ Next, define $\frac{T_{n-1}}{n!}=V_{n-1}$ to get $$V_{n}=\frac{1}{4} V_{n-1} \implies V_n=(\frac{1}{4})^n C \implies \frac{T_n}{(n+1)!}=\frac{C}{4^n},$$ giving back (1), where the constant $C=5$.