The way to think of this is that ## gets replaced by # in just the
same way that #1 gets replaced by ‘whatever is the first argument’.
So if you define a macro:
\newcommand\a[1]{+#1+#1+#1+}
or (using the TeX primitive \def):
\def\a#1{+#1+#1+#1+}
and use it as \a{b},
the macro expansion produces ‘+b+b+b+’,
as most people would expect.
However, if we now replace part of the macro:
\newcommand\a[1]{+#1+\newcommand\x[1]{xxx#1}}
then \a{b} will give us the rather odd
+b+so that the new\newcommand{x}[1]{xxxb}
\x ignores its argument.
If we use the TeX primitive:
\def\a#1{+#1+\def\x #1{xxx#1}}
\a{b} will expand to ‘+b+\def\x bb;xxxb}’. This
defines \x to be a macro delimited by b, and taking no
arguments, which is surely not what was intended!
Actually, to define \x to take an argument, we need
\newcommand\a[1]{+#1+\newcommand\x[1]{xxx##1}}
or, using the TeX primitive definition:
\def\a#1{+#1+\def\x ##1{xxx##1}}
and \a{b} will expand to
+b+\def\x #1{xxx#1}
because #1 gets replaced by ‘b’
and ## gets replaced by #.
To nest a definition inside a definition inside a definition then you
need ####1, doubling the number of # signs; and at the next level
you need 8 #s each time, and so on.
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