Solution:
Since $5^{2} = 25$ ends in $5$ again, and $6^{2} = 36$ ends in $6$ again, no matter to what power we raise $5$ or $6$, the resulting numbers will end in $5$ or $6$, respectively. Thus, $1995^{1995}$ ends in $5$ and $1996^{1996}$ ends in $6$.

Further, $4^{1} = 4$, $4^{2} = 16$, $4^{3} = 64$, $4^{4} = 256$, etc., i.e., an odd power of $4$ ends in $4$, and an even power of $4$ ends in $6$ (why?). Hence, $1994^{1994}$ ends in $6$.

So, the last digit of $1994^{1994} + 1995^{1995} + 1996^{1996}$ is the same as the last digit of $6 + 5 + 6 = 17$, i.e., it ends in $7$.

It remains to show that $1993^{1993}$ ends in $3$ (so that $3 + 7 = 10$).

Indeed, $3^{1} = 3$, $3^{2} = 9$, $3^{3} = 27$, $3^{4} = 81$, $3^{5} = 243$, and the pattern will be $3, 9, 7, 1, 3, 9, 7, 1, \ldots$, etc. So we need to figure out into which of these slots the $1993$rd power of $3$ will fall. Equivalently, we need to find the remainder of $1993$ when divided by $4$ (the length of the pattern above is $4$). Now, $1993 = 4 \cdot 498 + 1$: $3^{1993}$ will complete $498$ cycles of the pattern, and will end in $3$.

This shows that $1993^{1993}$ ends in $3$. Therefore, the last digit of the total sum is the last digit of $3 + 6 + 5 + 6 = 20$, $0$, which makes the sum divisible by $10$.