Proof: By rule of formation $$a_1=1=\frac{(1+1)!}{2}$$
$$a_2=3=\frac{(1+2)!}{2}$$
$$a_3=12=\frac{(1+3)!}{2}$$
$$a_4=60=\frac{(1+4)!}{2}$$
$$a_5=360=\frac{(1+5)!}{2}$$
$$a_6=2520=\frac{(1+6)!}{2}$$
$$\vdots$$
$$a_n=\frac{(1+n)!}{2}$$
Is there any other way?
Would appreciate your guidance.
On
When you have a sequence of integers like $1, 3, 12, 60, 360, 2520$ and you want to know what is a likely extension of it... consult the On-line Encyclopedia of Integer Sequences. Type your sequence in the search box, and see what happens!
In this case, we get 4 sequences that match it. But A00170 is the most likely. (The other three turned out to be essentially this one anyway.)
"Is there any other way?" Any other way for what? The n-th term is a constant times the factorial of $n+1,$ and the factorial function this has no exact closed-form formula that equals it. You could approximate it by Stirling's approximation, or other methods, but this will only be an approximation; it won't give you the exact value in the sequence.