The answer to this question is derived as follows. First we find the number of outcomes with all 6 different numbers on the die and then put it over the 6^10. 
I don't understand why we have to apply the principal of inclusion and exclusion here. Doesn't subtracting the second term, with the $5^{10}$, cover subtracting all the cases that we want to get rid of, because it only considers sequences with just 5 outcomes? I.e. Why do we add back the $4^{10}$ term?
On
Your particular query has been answered. However, {f you can use Stirling numbers of the second kind, which in your context would count 10 throws of the die with all 6 faces showing, the expression just becomes $$6!\cdot{10\brace6}$$
Where the second term represents the required Stirling number of the second kind
And by dividing it by $6^{10}$, we get
$Pr \approx 0.271812$
On
Without electronic tool, only paper+pencil.
We look at first die, nothing to say
We look at 2nd die, it can have same value (p=1/6) or new value (p=5/6)
We look at 3rd die. if the 2 first had same value, we can have still same value $(p=1/6 \times 1/6)$, or new value $(p=1/6 \times 5/6)$. If the 2 first values were different, we can have a value already in this set $(p=5/6 \times 2/6)$ or a new value $(p=5/6 \times 4/6)$
So, after 3 dice, we can have 1 value (p=1/36), 2 values (p=15/36) or 3 values (p=20/36)
We look at 4th die. If we have only 1 value until now, we can have still same value $(p=1/36 \times 1/6)$ or a new value $(p=1/36 \times 5/6)$. etc
After 4th die, p(1)=1/216, p(2)=35/216, p(3)=120/216, p(4)=60/216
After 5th die, p(1)=1/1296, p(2)=75/1296, p(3)=500/1296, p(4)=600/1296, p(5)=120/1296
After 6th die, p(1)=1/7776, p(2)=155/7776, p(3)=1800/7776, p(4)=3900/7776, p(5)=1800/7776, p(6)=120/7776
At this step, if we have only 1 value, the game is lost, we will not be able to reach 6 with the 4 remaining dice. And if we have already 6 different values, we win the game.
After 6th die, p(LOST)=1/7776, p(2)=155/7776, p(3)=1800/7776, p(4)=3900/7776, p(5)=1800/7776, p(WIN)=120/7776
If LOST or WIN, it is even not useful to look at remaining dice.
After 7th die, p(LOST)=1/7776, p(LOST=2)=310/(67776), p(3)=6330/(67776), p(4)=21000/(67776), p(5)=16800/(67776), p(WIN)=2520/(6*7776)
After 8th die, p(LOST)=1/7776, p(LOST=2)=310/(67776), p(LOST=3)=18990/(367776), p(4)=102990/(367776), p(5)=126000/(367776), p(WIN)=31920/(36*7776)
After 9th die, p(LOST)=1/7776, p(LOST=2)=310/(67776), p(LOST=3)=18990/(367776), p(LOST=4)=411960/(2167776), p(5)=835980/(2167776), p(WIN)=(631920+126000)/(2167776)
After 10th die, P(WIN)= 835980/(62167776) + (631920+126000)/(2167776)
If we look at numeric values, we see that proba of Win after 7° die is 5.4%, then 11.4% after 8° die, 18.9% after 9° die and 27.2% after 10° die. It seems totally consistant, and we will reach 100% very soon if we have more than 10 dice.
Edit : I see 27.18% in another answer, and I have 27.199% Probably I did some errors.
We add back the $4^{10}$ term because if we don't we end up overcounting the "bad" combinations we have subtracted. For example, any combination that omits both $1$ and $4$ will be counted twice in the $5^{10}$ term -- once because it omits $1$, and again because it omits $4$.
The principle of inclusion and exclusion accounts for all of the overcounting and undercounting caused by the presence of a combination in multiple subcases.