How many passwords are possible in this scenario? A password string consists of one or more of the 26 characters A..Z and can be of any length from $1$ to $8$ characters. How many password are possible in this scenario?

As my calculation, it should be  $2^{26}\times8!$, is it correct?
for one 1 char passwords, one can only put $26$ different characters
for one 2 char passwords, one can only put  $26^2$ different characters
for one 3 char passwords, one can only put  $26^3$ different characters
...
for one 8 char passwords, one can only put $26^8$ different characters

In short, think each box can contain how many letters from the given alphabet and multiply the values in the boxes. Finally, sum them.
$$\sum_{i=1}^8 26^i = \frac{26^9-1}{26-1}-1$$
The name of the formula is Sum of Consecutive powers of a number;
$$\sum_{i=0}^n a^i = \dfrac{a^{n+1} -1}{a-1}$$
and $-1$ since $i=0$ is not a case here.

Update : a more general case;
Let assume that you want to build a web application and want to determine the security level of the passwords. Here a calculator;

Assume that the alphabet has $l$ letters,
$m$ numerals, and
$p$ non-alphanumeric.

If we set the passwords to require at least;

$n_m$ numerals
$n_p$ non-alphanumeric,with
passords's length to $$min \geq n_k+n_a \geq 6,\text{and}$$
$max &gt; min$, then

What is the password space, $\mathcal{S}$?
$$ \mathcal{S} =  m^{n_m} + p^{n_p} + \sum_{i=1}^{max - ({n_m} + {n_p})} (l+m+p)^i$$
Under these assumptions, let

$l = 26$

$m = 10$

$p = 20$

$min = 6$

$max=20$

$n_m=2$

$n_p =1$, then we have;
$$ \mathcal{S}_{6:20} =  10^2 + 20 + \sum_{i=1}^{17} (56)^i = 533361663473057950558105648760 $$


$ \mathcal{S}_{6:20}$ has 99 digits in binary form.
if max is;

$6 \text{ then } S = 178928$
$7 \text{ then } S = 10013424$
$8 \text{ then } S = 10013424$
$9 \text{ then } S = 560745200$
$10\text{ then } S = 31401724656$ is 35-bit.

WolframAlpha calculator;
m^{n_m} + p^{n_p} + sum (l+m+p)^i, i=1 to (max - (n_m + n_p))