Try a proof by contradiction.
the idea is very simple. to "keep track", let's call the number of objects distributed in the i-th pigeonhole Mi.
first we check M1, by counting the objects in the first pigeonhole. if M1 ≥ n1, we are done, it is the pigeonhole the theorem claims exists.
but maybe not. so we move on to M2. if M2 ≥ n2, again, we are good, we can use M2 as the desired pigeonhole.
but let's say it's just the worst possible case, the first t-1 pigeonholes all have fewer objects than the corresponding ni.
how many objects do we have in all?
n1 + n2 +...+ nt - t + 1.
what's the MOST possible objects we could have come across in our first t-1 pigeonholes?
(n1 - 1) + (n2 - 1) +...+ (nt-1 - 1) = n1 + n2 +...+ nt-1 - (t-1).
this means that there is at LEAST:
[n1 + n2 +...+ nt - t + 1] - [n1 + n2 +...+ nt-1 - (t-1)] = nt objects in the last pigeonhole.
it's easier to see what is happening with a small number of pigeonholes.
suppose we have only one, and we have k objects distributed in 1 pigeonhole. then that pigeonhole has at least k objects (duh!).
perhaps that's "too easy". ok, let's try TWO pigeonholes, with n1 + n2 - 1 objects in 2 pigeonholes.
if the first pigeonhole has n1 or more objects, we're done. if not, it has at most n1 - 1. objects. that means there's at least n2 objects left over which have to be in the 2nd pigeonhole.
the same logic applies to 3 pigeonholes and n1 + n2 + n3 - 2 objects. rather than go through the same steps, let's pick 3 numbers, like 3,6, and 8. 3+6+8 = 17. 17 - 2 = 15.
so we have 15 objects to place in 3 pigeonholes. can we "beat the theorem" by placing less than:
3 objects in the first pigeonhole (ok, we want to get rid of "as many as we can" so we put 2. now we have 13 left).
6 objects in the 2nd pigeonhole (the most we can put is 5, right? that leaves us with...uh...8 left).
8 objects in the 3rd pigeonhole (well, shucks...we're stuck!)?
do you see why the theorem claims we have n1+..+nt - t + 1 = n1+..+nt - (t-1) objects?
if we had subtracted from the sum of the ni, t objects, it would be possible to put ni-1 objects in each pigeonhole, summing to a total of n1+...+nt - t objects.
by only subtracting t - 1 objects, we ensure at least ONE pigeonhole has ni objects (the "extra one we didn't subtract").