For simplicity, let us rename

as

for each

. Then we have the following equations:

So we must prove the above recursion satisfies:

. By induction on

.
If

the proposition holds trivially.
Now suppose that for all

,

. This is our induction hypothesis. In particular, we will asumme that it is true when

and
(*). We want to show the proposition also holds when

.
Claim:
Note that:

But

. So

as we want.
Alternatively, it is possible to give a combinatorial argument to demonstrate this formula based on

represents the number of desarrangements of length

.
Regards.
(*) Maybe it would have been more suitable to use a variant of the strong induction principle, which says:
Let
be a property, then
holds for all
if:i)-
ii)- 
Note that if we use this theorem it is necessary to check
two base cases istead of only one. ¿What is the advantage? Certainly there is no advantage, but (from my modest point of view) is more elegant if the in the inductive step we asumme no more hypothesis than strictly necessary.