All aboard the factorial train
Jelly, &nbsp;5&nbsp; 4 bytes

!:RS


Try it online! Or see the test-suite.

How?

When we introduce an \$n^{\text{th}}\$ train it allows:


\$(n-1)!\$ states - by being placed behind none of the \$n-1\$ existing trains and being faster than all of them.
all the previous end states, each in \$n\$ different ways - by being placed behind at least one existing train and being faster than \$[0,n-1]\$ of the \$n-1\$ existing trains


We know \$a(1) = 1\$ so...  

n  a(n) 
1  1
2  1! + 1*2
3  2! + 1!*3 + 1*2*3
4  3! + 2!*4 + 1!*3*4 + 1*2*3*4
5  4! + 3!*5 + 2!*4*5 + 1!*3*4*5 + 1*2*3*4*5
... etc


Which is:

n  a(n) 
1  1
2  1 + 2
3  1*2 + 1* 3 + 2*3
4  1*2*3 + 1*2* 4 + 1* 3*4 + 2*3*4
5  1*2*3*4 + 1*2*3* 5 + 1*2* 4*5 + 1* 3*4*5 + 2*3*4*5
...
n  n!/n + n!/(n-1) + n!/(n-2) + ... + n!/1


Hence the code:

!:RS - integer, n
!    - (n) factorial      n!
  R  - range (n)          [1,...,n-2,n-1,n]
 :   - integer division   [n!/1,...,n!/(n-2),n!/(n-1),n!/n]
   S - sum                a(n)