Write down the steps in the calculation of the fft for m = 2^4

The procedure is effectively identical for m = 2^4 as for any other
power of 2.  As described on page 348, you take the summation and
group terms by even and odd indices.  Then you factor out z^j from the
odd-index sum.

The general algorithm is fully described on pages 350-352, so I'm not
sure exactly what I'm supposed to say to answer this question.
Anyway, here are the (tedious) formulae representing all the matrix
transformations for this problem.  I hope this is the answer you were
looking for.

Pc
c0
c8
c1
c9
c2
c10
c3
c11
c4
c12
c5
c13
c6
c14
c7
c15

d = D1SPc
d0 = c0 + c8
d1 = c0 - c8
d2 = c1 + c9
d3 = z^2(c1 - c9)
d4 = c2 + c10
d5 = z^4(c2 - c10)
d6 = c3 + c11
d7 = z^6(c3 - c11)
d8 = c4 + c12
d9 = i(c4 - c12)
d10 = c5 + c13
d11 = z^10(c5 - c13)
d12 = c6 + c14
d13 = z^12(c6 - c14)
d14 = c7 + c15
d15 = z^14(c7 - c15)

e = D2SPd
e0 = d0 + d8
e1 = d0 - d8
e2 = d1 + d9
e3 = d1 - d9
e4 = d2 + d10
e5 = z^4(d2 - d10)
e6 = d3 + d11
e7 = z^4(d3 - d11)
e8 = d4 + d12
e9 = i(d4 - d12)
e10 = d5 + d13
e11 = i(d5 - d13)
e12 = d6 + d14
e13 = z^12(d6 - d14)
e14 = d7 + d15
e15 = z^12d7 - d15)

g = D3SPe
g0 = e0 + e8
g1 = e0 - e8
g2 = e1 + e9
g3 = e1 - e9
g4 = e2 + e10
g5 = e2 - e10
g6 = e3 + e11
g7 = e3 - e11
g8 = e4 + e12
g9 = i(e4 - e12)
g10 = e5 + e13
g11 = i(e5 - e13)
g12 = e6 + e14
g13 = i(e6 - e14)
g14 = e7 + e15
g15 = i(e7 - e15)

QTf = Sg
f0 = g0 + g8
f8 = g0 - g8
f4 = g1 + g9
f12 = g1 - g9
f2 = g2 + g10
f10 = g2 - g10
f6 = g3 + g11
f14 = g3 - g11
f1 = g4 + g12
f9 = g4 - g12
f3 = g5 + g13
f11 = g5 - g13
f5 = g6 + g14
f13 = g6 - g14
f7 = g7 + g15
f15 = g7 - g15