
Stirling and Bell Numbers 61
Proof (1) By Exercise 3.7, we have that
π∈Π
n
(x)
b
π
= x
n
,whereb
π
is the number of
blocks in π. Applying L to this equation, we obtain by linearity of L and by (3.21) that
L(x
n
)=
π∈Π
n
L((x)
b
π
)=
π∈Π
n
1=
n
,asclaimed.Toshow(2),wefirstobservethat
due to x(x − 1)
n
=(x)
n+1
we have L(x(x − 1)
n
)=L((x)
n+1
)=1=L((x)
n
), which,
by linearity of L, implies L(xp(x − 1)) = L(p(x)) for every polynomial p.Inparticular,
for p(x)=(1+x)
n
,weobtainL(x
n+1
)=L((1 + x)
n
)=
n
j=0
n
j
L(x
j
). Using (1), we
obtain (2). To show (3), we use the series expansion of e
x
and get e
x
=
j≥0
x
j
j!
. Taking
the nth derivative and evaluating at x =1gives1=
1
e
j≥0
(j)
n
j!
. By (3.21),