By induction, assume that
n
5

n
is divisible by 5 for
n
=
k
. Then, for
n
=
k
+ 1,
(
k
+ 1)
5

(
k
+ 1) =
k
5
+ 5
k
4
+ 10
k
3
+ 10
k
2
+ 5
k
+ 1

k

1
= (
k
5

k
) + 5(
k
4
+ 2
k
3
+ 2
k
2
+
k
)
By the inductive hypothesis,
k
5

k
is divisible by 5, and the second
term is divisible by 5 because it is 5 times an integer, so (
k
+1)
5

(
k
+1)
is divisible by 5. Therefore, by induction
n
5

n
is divisible by 5 for all
nonnegative integers
n
.
§
4.1, #50: For
n
= 1, the statement is that for a set of 1 + 1 = 2 positive integers,
none exceeding 2
·
1 = 2, there is at least one integer that divides
another.
But, the only possible such set is
{
1
,
2
}
, and 1 divides 2.
Therefore, the statement is true for
n
= 1.
For the inductive step, we assume that for every set of
n
+1 integers at
most 2
n
, there is an integer in the set dividing another in it. Let
S
be a
set of
n
+2 integers which are at most 2(
n
+1). Either
S
contains neither
2
n
+ 1 nor 2
n
+ 2,
S
contains one of them, or
S
contains both. In the
first case, for an arbitrary element
k
in
S
,
S
{
k
}
has
n
+1 elements at
most 2
n
, so by the inductive hypothesis, one of them divides another.
In the second case,
S
{
2
n
+1
,
2
n
+2
}
has the same property, so again,
by the inductive hypothesis, one integer in
S
divides another.