Discrete mathematics problem - Probability theory and counting -

I am taking discrete mathematics courses, and I had to face a question and I need your help. I do not know that this is the right place for here :)

It says:

There is a password of every user on the computer system, which is six to eight characters long, where Each letter should have an uppercase letter or a digit, each password must have at least one digit. How many possible passwords are there?

This book solves it by adding probabilities of long passwords up to six, seven and eight characters, however, when it resolves the possibility of six characters, then it does

 P6 = 36  6  - 26  6 and 
 
 P7 = 36  7  - 26  7  

and

 P8 = 36  8  - 26  8  

And then add them all.

I understand the solution, but my question is, why not, P6 = 10 * 36 ⁵ and work for P7 and P8? 10 for points and 36 for alphanumeric?

Besides, if someone can give me some other solution in the book.

Thank you very much :)

You forget that the number is in any case May be.