No, Inclusion-Exclusion is not that simple. See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
For any set $~E~$ with a finite number of elements, let $~|E|~$ denote the number of elements in the set $~E.$
Let $~S~$ denote all of the passwords possible, if the requirements of digits, special characters, upper case letters, and lower case letters are ignored.
Let $~S_1~$ denote the subset of $~S~$ that violates the constraint that at least one digit must be present.
Let $~S_2~$ denote the subset of $~S~$ that violates the constraint that at least one special character must be present.
Let $~S_3~$ denote the subset of $~S~$ that violates the constraint that at least one lower case letter must be present.
Let $~S_4~$ denote the subset of $~S~$ that violates the constraint that at least one upper case letter must be present.
Then, the desired computation is
$$|S| - |S_1 \cup S_2 \cup S_3 \cup S_4|. \tag1 $$
Let $~T_0~$ denote $~|S|.$
For $~r \in \{1,2,3,4\},$
let $~T_r~$ denote $\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 4} | ~S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r} ~|.$
That is, $~T_r~$ denotes the sum of $~\displaystyle \binom{4}{r}~$ terms.
Then, in accordance with Inclusion-Exclusion theory,
the computation in (1) above equals
$$\sum_{r=0}^4 (-1)^r T_r. \tag2 $$
So, the problem reduces to computing
$~T_0, ~T_1, ~T_2, ~T_3, ~T_4.$
$\underline{\text{Computation of} ~T_0}$
There are $~77~$ choices for each of $~8~$ letter positions.
Therefore,
$$T_0 = (77)^8.$$
$\underline{\text{Computation of} ~T_1}$
To compute $~|S_1|~$ note that there are $~67~$ choices for each of the $~8~$ character positions. This is because $~S_1~$ represents the subset of $~S~$ where all digits are excluded. This means that instead of there being $~77~$ choices for each character position, there are only $~(77 - 10) = 67~$ choices for each character position.
Therefore, $~|S_1| = (67)^8.$
Very similarly:
$~|S_2| = (62)^8.$
$~|S_3| = (51)^8.$
$~|S_4| = (51)^8.$
Therefore,
$$T(1) = |S_1| + |S_2| + |S_3| + |S_4| \\
= (67)^8 + (62)^8 + (51)^8 + (51)^8.$$
$\underline{\text{Computation of} ~T_2}$
Similar to the analysis in the previous section,
to compute $|S_1 \cap S_2|,~$ you must reason that with no digits or special characters, there are $~52~$ choices for each of the 8 character positions.
Therefore, $~|S_1 \cap S_2| = (52)^8.$
Similarly:
$~|S_1 \cap S_3| = (41)^8.$
$~|S_1 \cap S_4| = (41)^8.$
$~|S_2 \cap S_3| = (36)^8.$
$~|S_2 \cap S_4| = (36)^8.$
$~|S_3 \cap S_4| = (25)^8.$
Therefore,
$$T_2 = (52)^8 + (41)^8 + (41)^8 + (36)^8 + (36)^8 + (25)^8.$$
$\underline{\text{Computation of} ~T_3}$
$|S_1 \cap S_2 \cap S_3| = (26)^8,~$
since this represents the set where only capital letters are used in each of the $~8~$ character positions.
Similarly:
$|S_1 \cap S_2 \cap S_4| = (26)^8.$
$|S_1 \cap S_3 \cap S_4| = (15)^8.$
$|S_2 \cap S_3 \cap S_4| = (10)^8.$
Therefore,
$$T_3 = (26)^8 + (26)^8 + (15)^8 + (10)^8.$$
$\underline{\text{Computation of} ~T_4}$
$S_1 \cap S_2 \cap S_3 \cap S_4 ~$ represents the set of all 8 character sequences with no digits, special characters, lower case letters or upper case letters.
Therefore, $~S_1 \cap S_2 \cap S_3 \cap S_4 = \emptyset.$
Therefore,
$$T_4 = 0.$$
$\underline{\text{Final Computation}}$
$$T_0 = (77)^8.$$
$$T(1) = (67)^8 + (62)^8 + (51)^8 + (51)^8.$$
$$T_2 = (52)^8 + (41)^8 + (41)^8 + (36)^8 + (36)^8 + (25)^8.$$
$$T_3 = (26)^8 + (26)^8 + (15)^8 + (10)^8.$$
$$T_4 = 0.$$
The expression in (1) above equals
$$T_0 - T_1 + T_2 - T_3 + T_4.$$
