Prove that there is $[e ~(b-1) ~(b-1)!]$ natural numbers with no repeating digits in base $b$ For example, in base $2$ we have exactly $2$ of them (not counting zero):
$$1,~10$$
In base $3$ we have $10$ (if I'm correct):
$$1,2,10,12,20,21,102,120,201,210$$
By observation of these simple cases I see that the amount of such numbers is:
$$N_b=(b-1)+(b-1)^2+(b-1)^2(b-2)+(b-1)^2(b-2)(b-3)+\dots+ \\ +(b-1)^2(b-2)\cdots(b-(b-1))$$
So we have:
$$N_4=48$$
$$N_5=260$$
$$N_{10}=8877690$$

Is this formula correct? Does it have a closed form (or a more convenient at least)?

Edit
I found the sequence here: http://oeis.org/A036918
The formula is:
$$N_b=[e ~(b-1) ~(b-1)!]$$

How to prove it?

 A: For an $m$-tuple of distinct base-$b$ digits, $1 \le m \le b$, there are
$\dfrac{b!}{(b-m)!}$ possibilities.  For a number of $m$ digits, with leading $0$ not allowed, there are $$a_m = \dfrac{b-1}{b} \dfrac{b!}{(b-m)!}$$
In total we have
$$ \sum_{m=1}^b a_m = \dfrac{b-1}{b} \sum_{j=0}^{b-1} \dfrac{b!}{j!}
= (b-1)\sum_{j=0}^{b-1} \dfrac{(b-1)!}{j!}$$
For convenience, write $b-1 = c$.   $$c!\; e = \sum_{j=0}^\infty \dfrac{c!}{j!}$$
The terms for $j \le c$ are integers, while for $j > c$ we have
$$ \sum_{j=c+1}^\infty \dfrac{c!}{j!} < \sum_{k=1}^\infty \dfrac{1}{(c+1)^{k}} = \dfrac{1}{c} $$ 
Thus 
$$ c \sum_{j=0}^c  \dfrac{c!}{j!} < c! \; c e < 1 + c \sum_{j=0}^c  \dfrac{c!}{j!} $$
i.e. $$\lfloor c!\; ce \rfloor = c \sum_{j=0}^c \dfrac{c!}{j!} = \sum_{m=1}^b a_m$$
A: We can determine this directly by writing out the number of . In particular, the number of strings of distinct numbers in $\{0,1,\ldots,b-1\}$ of length $n$ not starting with $0$ is easily seen as $$\frac{b-1}b\cdot \frac{b!}{(b-n)!}$$
which, when expanded, is the same thing that you wrote. If we sum this, we get 
$$N_k=\frac{b-1}b\sum_{n=1}^{b}\frac{b!}{(b-n)!}=(b-1)\cdot (b-1)!\cdot \sum_{n=1}^b\frac{1}{(b-n)!}$$
Going a little further, we may rewrite the sum by substitution giving:
$$N_k=(b-1)\cdot (b-1)! \cdot \sum_{u=0}^{b-1}\frac{1}{u!}.$$
Now, we just note that, if we let $b-1$ go to infinity, the sum alone would rapidly converges to $e$. In particular, consider the difference
$$e(b-1)(b-1)!-N_k=\sum_{u=b}^{\infty}\frac{(b-1)(b-1)!}{u!}.$$
If we can show that this difference is less than $1$, we're done. To do so, note that $u!>b!\cdot b^{u-b}$ for $u>b$. Using this bound:
$$e(b-1)(b-1)!-N_k<\sum_{u=b}^{\infty}\frac{(b-1)(b-1)!}{b!\cdot b^{u-b}} = \frac{b-1}b\cdot \sum_{v=0}^{\infty}\frac{1}{b^v}=\frac{b-1}b\cdot \frac{b}{b-1}=1$$
So we are done.
A: There is a recurrence formula for $N_k$.
First of all rewrite $N_k$ as
$$
N_k = (k-1)\Big[1 + (k-1) + (k-1)(k-2) + \dots + (k-1)!\Big].
$$
Now it's easy to see that
$$
N_{k+1} = \left(\frac{N_1}{k-1}k + 1\right)k = \frac{k^2}{k-1}N_k + k.
$$
For example $N_3 = \frac{2^2}{1}N_2 + 2 = 10$, $N_4 = \frac{3^2}{2}N_3 + 3 = 48$.

A bit deeper
We may substitute in that rucurrent relation $N_k = \frac{(k-1)^2}{k-2}N_{k-1} + k - 1$ and get
$$
N_{k+1} = \frac{k^2(k-1)}{k-2}N_{k-1} + k(k-1). 
$$
Thus 
$$
N_{k+1} = (1-\alpha)\left(\frac{k^2}{k-1}N_k + k\right) + \alpha\left(\frac{k^2(k-1)}{k-2}N_{k-1} + k(k-1)\right),
$$
and we may take $\alpha$ such that $(1-\alpha)k + \alpha k(k-1) = 0$, i.e. $\alpha = -1/k$. Thus
$$
N_{k+1} = \frac{(k+1)k}{k-1}N_k - \frac{k(k-1)}{k-2}N_{k-2}.
$$
For example $N_4 = \frac{4\cdot 3}{2}N_3 - \frac{3\cdot 2}{1} N_2 = 48$.
Now let vector $\mathbf{N}_k = (N_k, N_{k-1})^{\top}$. We know that $\mathbf{N}_2 = (2,10)^\top$. If we denote matrix $A_k$ as
$$
A_k = \begin{pmatrix}
\dfrac{(k+1)k}{k-1} & -\dfrac{k(k-1)}{k-2} \\
1 & 0
\end{pmatrix}
$$
then we get
$$
\mathbf{N}_{k+1} = A_k\mathbf{N}_k 
$$
thus
$$
\mathbf{N}_n = \left(\prod_{k=3}^{n}A_k\right)\mathbf{N}_2.
$$
