Can you think of any shortcuts here so that you don’t have to manually compute each and every term’s value?

Do you need to actually compute each full $k$ term? Does it matter?

Since we need to find the units digit in base 10 of the sum $\displaystyle{ \sum_{k=1}^{99} k! }$, we are really trying to find $\displaystyle{ \left( \sum_{k=1}^{99} k! \right) \mod 10 }$.

First, we note that only the units digit of each term $k!$ has an influence on the units digit of the sum; in other words, we can consider just summing up each term’s units digit and then take the final digit of the sum:

$$ \left( \sum_{k=1}^{99} k! \right) \mod 10 = \left( \sum_{k=1}^{99} ( k! \mod 10 ) \right) \mod 10 $$

The reason this is valuable is that we don’t want to compute each $k!$ term, but we can find that many terms are $0 \mod 10$ because they include $10$ as a factor, so we can just ignore them entirely.

Since $k! = k \times (k-1) \times \cdots \times 1$, any term $k \ge 10$ can be excluded since $10 \cdot x \equiv 0 \mod 10$ for any $x$. Thus, we can exclude all terms in the interval $[10, 99]$ and can just compute the units digit of the terms in the interval $[1, 9]$.

However, we can save ourselves even more effort because $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \equiv 0 \mod 10$ because it includes both terms $5$ and $2$. Thus, we can further exclude all terms $k \ge 5$, so we eliminate the interval $[5, 9]$ as well, and we are just left with:

Thus, the units digit of the overall sum is $\boxed{ 3 }. \enspace \blacksquare$

Since Python supports arbitrarily-sized integers, we can easily verify this directly:

$ python
>>> import math
>>> sum([math.factorial(k) for k in range(1, 100)]) % 10
3

Note that Python’s range(x, y) function creates a half-open interval $[x, y)$:

>>> list(range(1, 10))
[1, 2, 3, 4, 5, 6, 7, 8, 9]

Thus, range(1, 100) is what we need to represent $\displaystyle{\sum_{k = 1}^{99}}$ above.