We will show that k=39. Namely, it is possible to make postage of every value
n > 39, but it is not possible to make a postage of 39.

Informally the idea is very simple: One can show that all values
40,41,42,43,44 can be obtained, and then every larger value can be
obtained by one of the above by adding an appropriate amount 5c
stamps. Formalizing this, we use complete induction:

P(n) = "there are natural numbers a nd b such that n = 5a+11b"
Let n be anumber bigger than 39.

Induction hypothesis "for all values m, m>39 and m<n P(m)"
We need to show for n that P(n) holds.
P(40) holds: 40 = 8 X 5 + 0 X 11
P(41) holds: 41 = 6 X 5 + 1 X 11
P(42) holds: 42 = 4 X 5 + 2 X 11
P(43) holds: 43 = 2 X 5 + 3 X 11
P(44) holds: 41 = 0 X 5 + 4 X 11
For n > 44 by induction hypotheis (notice n-5 > 39) we have that
there are a,b>=0 such that n-5 = 5a+11b. Then if we take a' = a+1, b'=b
we can easily verify that n = 5a' + 11b.

To show that 39c can not be obtained by 5c and 11c, simply use
elimination.  There can be at most 3 stamps of 11c in use, since
otherwise we would have at least 44c of postage. So for each of the
possible values b in {0,1,2,3} we can easily check (for example, with
a proof by cases) that the remaining amount that should come from the
5c stamps, namely 39 - 11b, is not divisible by 5, hence this is
impossible. (You get 39, 28, 17, 6 for b = 0,1,2,3,4, respectively.)