Grandparent: 5 random lower case characters + one upper case = 26^6 * 6 NOT 52 ^ 6
Parent: 5 random lower case characters + one upper case = 52^6. It would be 26^6 if and only if you knew exactly where the upper case letter was, which is an unreasonable assumption. Adding an upper case letter would eliminate a straight lower-case dictionary attack entirely and double the pool of possible characters from 26 to 52. There are 6 places, so 52^6.
The grandparent poster has done the calculation correctly, if it is assumed that the cracker knows that there is exactly one uppercase character.
We're all agreed that if there is a 6-letter all-lower-case password, there are 26^6 possible passwords (26 possible character choices in each of six positions), right? For five lower case letters and one upper case letter, we draw five lower case letters (26^5 possibilities) and one upper case letter (26^1 possibilities, because it can't be a lower case letter), and we have 6 choices as to where in the password we place the upper case letter: 26^5 * 26^1 * 6 = 26^6 * 6 possible passwords.
Alternatively, consider our six-letter all-lower-case password and its 26^6 possibilities. We have a dictionary that starts aaaaaa, aaaaab, aaaaac and ends with zzzzzz. If we add exactly one (no more, no fewer) capital letter, then each entry in our original dictionary is replaced by six new passwords, one with a single capital letter in each position: Aaaaaa, aAaaaa, aaAaaa, aaaAaa, aaaaAa, aaaaaA, then Aaaaab, aAaaab, aaAaab, aaaAab, aaaaAb, aaaaaB, and so forth--again giving us 26^6 * 6 possible passwords.
That said, it would be unusual for our hypothetical cracker to have access to that sort of specific information about a password. Why would he know that there was exactly one upper case letter? Far more likely would be some sort of rudimentary password screen that required our password to contain a mix of capital and lower case letters--that is, at least one upper case, and at least one lower case. In that more-likely scenario, the parent's calculation is closer to the mark. Each of six positions could have any one of 52 values (26 upper- and 26 lower-case letters), giving 52^6 possibilities, from which we subtract 2*26^6 options, representing the forbidden all-lower-case and all-caps passwords, leaving 52^6-2*26^6 possible choices.