# Overlapping vertical spacing in math mode

You wrote,

If a better method exists to only get the equation tag on the last line ..., please let me know.

I think you need a combination of an equation environment, an aligned[b] environment, and a \smash[t]{...} directive.

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\label{eq.14} \begin{aligned}[b] % 'b' means "place equation number on bottom line" 1 \uparrow 1 &= {}^1 1 = 1 \\ 2 \uparrow\uparrow 2 &= {}^2 2 = 4 \\ 3 \uparrow\uparrow \uparrow 3 &= \smash[t]{ {}^{{}^{{}^3\!}} {}^3 3 = 3 \uparrow\uparrow 3 \uparrow\uparrow 3 = \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{3^{3^3}\text{ threes}} } \end{aligned}
\end{document}


Addendum: Here's an additional solution, which (a) employs an alignedat environment to perform alignment on both the first and second instance of =, (b) changes the math status of \uparrow, \uparrow\uparrow, and \uparrow\uparrow\uparrow to math-bin (the default math status is math-rel), and (c) implements @egreg's suggestion to provide a bit more distance between the rows and to enlarge the contents of the _{...} component of the \underbrace directive in row 3.

\documentclass{article}
\usepackage{mathtools}

\begin{document}
\label{eq.14} \addtolength\jot{4pt} \begin{alignedat}[b]{2} % 'b' means "place equation number on bottom line" 1 \mathbin{\uparrow} 1 &= {}^1 1 &&= 1 \\ 2 \mathbin{\uparrow\uparrow} 2 &= {}^2 2 &&= 4 \\ 3 \mathbin{\uparrow\uparrow\uparrow} 3 &= \smash[t]{ {}^{{}^{{}^3\!}} {}^3 3} &&= \smash[t]{3 \mathbin{\uparrow\uparrow} 3 \mathbin{\uparrow\uparrow} 3 = \underbrace{3^{3^{3^{3^{3^{3^{.^{.^{.^{3}}}}}}}}}}_{\mathclap{\textstyle 3^{3^3}\text{ threes}}} } \end{alignedat}
\end{document}


The main idea is the same as Mico's about smashing the top of the big object, but with some improvements on spacing and size of characters.

\documentclass{article}
\usepackage{amsmath}

\ExplSyntaxOn
\NewDocumentCommand{\hyop}{m}
{
\mathbin{ \prg_replicate:nn { #1 } { \uparrow } }
}
\ExplSyntaxOff
\newcommand{\preexp}[2]{\vphantom{#2}^{#1}\kern-\scriptspace#2}

\begin{document}
\label{eq.14} \begin{aligned}[b] % 'b' means "place equation number on bottom line" 1 \hyop{1} 1 &= \preexp{1}{1} = 1 \\ 2 \hyop{2} 2 &= \preexp{2}{2} = 4 \vphantom{\preexp{\preexp{3}{3}}{3}}\\ 3 \hyop{3} 3 &= \preexp{\preexp{3}{3}}{3} = 3 \hyop{2} ( 3 \hyop{2} 3 ) = \smash[t]{ \underbrace{ 3^{3^{3^{3^{3^{3^{{\cdot}^{{\cdot}^{{\cdot}^{3}}}}}}}}} }_{\text{\displaystyle 3^{3^3} threes}} } \end{aligned}

\end{document}