It's all about the Bonus Points

$\begin{aligned}B_n&=nB_{n-1}+1\\&=n((n-1)B_{n-2}+1)+1\\&=n(n-1)((n-2)B_{n-3}+1)+n\\&\cdots\\&\cdots\\&=n!B_0+1+n+n(n-1)+\cdots +n!\\&\large{=\sum_{r=0}^{n}\dfrac{n!}{r!}}~~\small{(\because ~B_0=1)}\end{aligned}$ $\large B_n=\sum_{r=0}^{n}\dfrac{n!}{r!}$ For $n=10$ . $\large B_{10}=\sum_{r=0}^{10}\dfrac{10!}{r!}=\boxed{9864101}$

As Rishabh said,

$B_n = nB_{n-1}+1 \\ \;\;\;\;\,= n((n-1)B_{n-2}+1)+1 = n(n-1)B_{n-2}+n +1 \\ \;\;\;\;\,= n(n-1)((n-2)B_{n-3}+1)+n +1 = n(n-1)(n-2)B_{n-3}+n(n-1)+n +1$

And now that we have the first few expressions, we should be able to see a pattern. If we continue the pattern until we have an expression for $B_n$ in terms of $B_0$ , that first term will in fact turn into n!. As for the other terms, set aside the 1 for the moment. The others that are created form a sequence: n, n(n-1), n(n-1)(n-2), etc. Another way to express this sequence is $\frac{n!}{(n-1)!}$ , $\frac{n!}{(n-2)!}$ , $\frac{n!}{(n-3)!}$ ..... Once again, take this sequence out as far as it will go, until the expression for $B_n$ is in terms of $B_0$ . We wind up with the term n(n-1)(n-2)(n-3)......4*3*2= $\frac{n!}{1!}$ . But remember, we're adding up all the terms; so to recap, we are looking at the formula for $B_n$ that is at the end of the sequence of formulae above. It has terms n!, $\frac{n!}{(n-1)!}$ , $\frac{n!}{(n-2)!}$ , $\frac{n!}{(n-3)!}$ , all the way down to $\frac{n!}{1!}$ , and 1. Now we can rewrite n! and 1 as $\frac{n!}{0!}$ and $\frac{n!}{n!}$ , respectively. Now we have a completely consistent pattern:

$B_n = \sum\limits_{i=0}^{n}{\frac{n!}{i!}}$

Now this is a result certainly, and it's a better than the recurrence relation, but it's still a summation. Can we eliminate it? Well the first thing is to factor out the n! to get

$B_n = n!\sum\limits_{i=0}^{n}{\frac{1}{i!}}$

Now we recall the Taylor Series definition of e:

$e = \sum\limits_{i=0}^{\infty}{\frac{1}{i!}}$

This suggests that

$B_n \approx n!*e$

Now about that "approximately" symbol. A few test cases suggests that we can turn it into a hard equals with the floor function:

$B_n = \lfloor n!*e \rfloor$

The last step is to prove that using the floor function won't ever stab us in the back. How do we know that adding up ALL the terms in the Taylor Series won't ever overwhelm the floor function? Using the full Taylor Series causes us to overestimate, and the floor function only works correctly if the amount of the overestimate is never greater than 1, so we need to prove that in fact, it never is. Take a closer look at the Taylor series expansion of n!*e:

$\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + ... + \frac{n!}{(n-2)!} + \frac{n!}{(n-1)!} + \frac{n!}{n!} +\\ \frac{n!}{(n+1)!} + \frac{n!}{(n+2)!} + \frac{n!}{(n+3)!} + ...$

I have broken it out into 2 lines; the first one matches our initial sum which we know is a precise expression for $B_n$ , and the second one is "the rest of" n!*e. We think, hope, that taking the floor function of this expression for n!*e will neatly remove the second line. For that to happen, the second line must by less than 1 always. Let's simplify it:

$\frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + ...$

It will suffice to show that:

1) As n increases, this expression always decreases, and
2) For some (small) n, it is less than 1.
To prove (1), define 2 consecutive elements of the sequence of series as

$D_n\;\;\, = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + ... \\ D_{n+1}= \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + ...$

This makes it clear that $D_{n+1}$ > $D_n$ , since the $n^{th}$ term of $D_{n+1}$ is less than the $n^{th}$ term of $D_n$ for all n.

For (2), we recognize that $D_{-1}=e$ . Also, the first 2 terms of e sum to 2, which is the whole part of e $\approx$ 2.71828. This allows us to conclude that $D_1 = e-2 \approx .71828..... < 1$ .

So we have proven that $D_n > 1$ for $n \ge 1$ , meaning that $B_n= \lfloor e*n! \rfloor$ for $n \ge 1$

One interpretation of the sequence can be given here although I cannot see how to take the sequence and derive the formula without having seen it before! Any insight is welcome :)

The simplest way of deriving the closed formula is to divide the equation by $n!$ , obtaining $\frac{B_n}{n!} \; = \; \frac{B_{n-1}}{(n-1)!} + \frac{1}{n!}$ It is now a simple induction to solve this recurrence relation for $\frac{B_n}{n!}$ , obtaining $\frac{B_n}{n!} \; = \; \sum_{j=0}^n \frac{1}{j!}$ so that $B_n \; = \; \sum_{j=0}^n \frac{n!}{j!}$ and $\boxed{B_{10} = 9864101}$ .