h(n) = n +1
h(h(n)) = h(n)+1 = n+2
h(h(h(h......n h's....h(n)))..)) = n+n = 2n
g(n)a = h(h(h........a h's...h(h(n))..)) = n+a
g(n)n = 2n
g(g(n)n)n = 3n
g(g(g(n)n)n)n = 4n
g(g(g(g......g(g(n)n)n....)n)n with n g's = n^2
f(n)a = g(g(g.......a g's........g(n)n)n...)n)n = n*a
f(n)n = g(g(g(g......g(g(n)n)n....)n)n with n g's = n*n = n^2
f(f(n)n)n = n^3
e(n)n = f(f(f(f......n f's......f(f(n)n)n....n)n)n = n^n = n^^2
e(e(n)n)n = n^^3
d(n)n = e(e(e....n e's...e(n)n...n)n = n^^n = n^^^2
d(d(n)n)n = n^^^3
overall, for any letter, it's just the letter after it nested. i.e: c(n)n = the same nesting to d(d...d(n)n)..)n as the other ones
h(n) = {8}(n)n
{7}(n)n= g(n)n
{M}(n)n is just the m-th letter of the alphabet with those two symbols after. The earlier the letter, the bigger it is.
When we get to 0, we have no letters for that, so we could just denote it as {0}(n)n and will follow the same pattern.
{8-n}(n)n = n{n}n or n^^^...n ^'s...^^^n, or n{n+1}n.
{8-({8-n}(n)n))}} = n{n{n+1}n}n (i think??)
and from there we can nest the letters infinitely, leading up to n{n{n.....n n's...n{n{n+1}n}n....n}n}n.
might make a part two some time but idk im extremely lazy lol