Simple recursion in LaTeX

\documentclass[border=15pt]{standalone}
\makeatletter
\def\tower{\@ifnextchar[{\def\endtower{}\towerstep}{}}%
\def\towerstep[#1]{#1%
  \@ifnextchar[{\edef\endtower{\endtower\egroup}^\bgroup\towerstep}{\endtower}}
\makeatother
\begin{document}
        $\tower[2][3][4][2]$ 
        $\tower[2][3][4]$ 
        $\tower[2][3]$ 
        $\tower[2]$ 
        $\tower$ 

\end{document}

A solution in the spirit of the programmation by continuation:

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand \towerAux { m o }
  {
    \IfValueTF { #2 }
      { \towerAux {{#1}{#2}} }
      { \__tower:nn {#1} { } }
  }

\cs_set:Npn \__tower:nn #1 #2 
  {
    \tl_if_empty:nTF { #1 }
      { #2 }
      { \__tower_i:nnn #1 { #2 } }
  }

\cs_set:Npn \__tower_i:nnn #1 #2 #3 { \__tower:nn { #1 } { #2 ^ { #3 } } }

\NewDocumentCommand \tower { } { \towerAux { } }
\ExplSyntaxOff

\begin{document}
$\tower[5][3][4][2]$
\end{document}

With this programmation, $\tower[5][3][4][2]$ is replaced successively by the following instructions:

$\tower[5][3][4][2]$

$\towerAux{}[5][3][4][2]$

$\towerAux{{}{5}}[3][4][2]$

$\towerAux{{{}{5}}{3}}[4][2]$

$\towerAux{{{{}{5}}{3}}{4}}[2]$

$\towerAux{{{{{}{5}}{3}}{4}}{2}}$

As you see, \towerAux is recursive.

Now, all the arguments (if I can say) have been structured in a kind of list and the last one is the first accessible. You can now construct the required result in a kind of auxiliary argument (at the end) as usual in recursive programmation. The commands \__tower:nn and \__tower_i:nnn are mutually recursive.

$\__tower:nn{{{{{}{5}}{3}}{4}}{2}}{}$

$\__tower_i:nnn{{{{}{5}}{3}}{4}}{2}{}$

$\__tower:nn{{{{}{5}}{3}}{4}}{2^{}}$

$\__tower_i:nnn{{{}{5}}{3}}{4}{2^{}}$

$\__tower:nn{{{}{5}}{3}}{4^{2^{}}}$

$\__tower_i:nnn{{}{5}}{3}{4^{2^{}}}$

$\__tower:nn{{}{5}}{3^{4^{2^{}}}}$

$\__tower_i:nnn{}{5}{3^{4^{2^{}}}}$

$\__tower_i:nnn{}{5^{3^{4^{2^{}}}}}$

$5^{3^{4^{2^{}}}}$

I build two token lists, the first containing

{1^{2^{3^{4^{5^{6^{7

and the other containing

}}}}}}}

Actually, the braces are stored as \c_group_begin_token and \c_group_end_token, so the token lists are balanced.

If [ follows, a further step is taken. At the end, the two lists are delivered.

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\tower}{}
 {
  \tl_clear:N \l__perner_tower_left_tl
  \tl_clear:N \l__perner_tower_right_tl
  \perner_tower_build:n { }
 }

\tl_new:N \l__perner_tower_left_tl
\tl_new:N \l__perner_tower_right_tl

\cs_new_protected:Nn \perner_tower_build:n
 {
  \peek_charcode:NTF [
   {% there is a [
    \__perner_tower_add:nw { #1 }
   }
   {% no [, end
    \__perner_tower_end:
   }
 }
    
\cs_new_protected:Npn \__perner_tower_add:nw #1 [#2]
 {
  \tl_put_right:Nn \l__perner_tower_left_tl { #1 \c_group_begin_token #2 }
  \tl_put_right:Nn \l__perner_tower_right_tl { \c_group_end_token }
  \perner_tower_build:n { \c_math_superscript_token }
 }

\cs_new_protected:Npn \__perner_tower_end:
 {
  \tl_use:N \l__perner_tower_left_tl
  \tl_use:N \l__perner_tower_right_tl
 }
\ExplSyntaxOff

\begin{document}

$\tower[1][2][3][4][5][6][7]$

\end{document}

Much shorter with a different syntax. The argument is split at commas; then between any two items we output ^{ (again as implicit tokens) and at the end the right number of } is output.

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\tower}{m}
 {
  \perner_tower_build:n { #1 }
 }

\seq_new:N \l__perner_tower_seq

\cs_new_protected:Nn \perner_tower_build:n
 {
  \seq_set_split:Nnn \l__perner_tower_seq { , } { #1 }
  \seq_use:Nn \l__perner_tower_seq { \c_math_superscript_token \c_group_begin_token }
  \prg_replicate:nn { \seq_count:N \l__perner_tower_seq - 1 } { \c_group_end_token }
 }
\ExplSyntaxOff

\begin{document}

$\tower{1,2,3,4,5,6,7}$

\end{document}