Quantum Harmonic Oscillator propagator in Sakurai In Sakurai the derivation of the propagator leads to the expression
$$u_n(x)\exp{\left(\frac{-iE_nt}{\hbar}\right)} = \left(\frac{1}{2^{n/2}\sqrt{n!}}\right) \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp{\left(\frac{-m\omega x^2}{2\hbar}\right)} H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)\exp{\left(-i\omega(n+\frac{1}{2})t\right)}$$
Which it says leads to the expression for the propagator by using the formula
$$\frac{1}{\sqrt{1-\zeta^2}}\exp{\left(\frac{-\xi^2-\eta^2+2\xi\eta\zeta}{(1-\zeta^2)}\right)} = \exp{\left[-(\xi^2+\eta^2)\right]\sum_{n=0}\left(\frac{\xi^n}{2^nn!}\right)H_n(\xi)H_n(\eta)}.$$
I’m wondering, what exactly is the reasoning for using this formula? Or how exactly does the transformation from this formula to the propagator start? Am I setting the prefactor equal to $\frac{1}{\sqrt{1-\zeta^2}}$ and solving for $\zeta$? For the inner product $\langle x’’ |\psi|x’\rangle$ Am I calling $\xi=x’’$ and $\eta=x’$? Any help would be much appreciated! 
EDIT: Formula has been fixed. Same questions as before remain
 A: You can find in Sakurai this expression for the propagator
$$K({\bf{x''}},t;{\bf{x'}},t) = \sum\limits_{a'} {\left\langle {x''\left| {a'} \right\rangle } \right.} \left\langle {a'\left| {x'} \right\rangle } \right.\exp \left[ {\frac{{ - i{E_{a'}}(t - {t_0})}}{\hbar }} \right]$$
Taking into account the QHO wavefunctions 
$${a_n}(x) = \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega {x^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x} \right),\qquad n = 0,1,2, \ldots .
$$
and
$$\left\langle {x''\left| {a'} \right\rangle } \right. = \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega x'{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right),
$$
We simply can derive the expression for $K({\bf{x''}},t;{\bf{x'}},t)$ 
$$\begin{array}{l}K({\bf{x''}},t;{\bf{x'}},t) = \sum\limits_{a'} {\frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {{\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)}^{1/4}} \cdot {e^{ - \frac{{m\omega x'{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)}  \times \\ \times \frac{1}{{\sqrt {{2^n}{\mkern 1mu} n!} }} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/4}} \cdot {e^{ - \frac{{m\omega x{'^2}}}{{2\hbar }}}} \cdot {H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right)\exp \left[ {\frac{{ - i{E_{a'}}(t - {t_0})}}{\hbar }} \right] = \\\frac{1}{{{2^n}{\mkern 1mu} n!}} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\sum\limits_{n = 0} {{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\exp \left[ { - i\omega (n + 1/2)(t - {t_0})} \right]}  = \\ = \frac{1}{{{2^n}{\mkern 1mu} n!}} \cdot {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\sum\limits_{n = 0} {{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\exp \left[ { - i\omega (n + 1/2)(t - {t_0})} \right]}  = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}{\sum\limits_{n = 0} {\frac{1}{{{2^n}{\mkern 1mu} n!}}{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\left( {{e^{ - i\omega (t - {t_0})}}} \right)} ^n} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}{e^{ - \frac{{m\omega (x{'^2} + x'{'^2})}}{\hbar }}}{\sum\limits_{n = 0} {\frac{1}{{{2^n}{\mkern 1mu} n!}}{H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x'} \right){H_n}\left( {\sqrt {\frac{{m\omega }}{\hbar }} x''} \right)\left( {{e^{ - i\omega (t - {t_0})}}} \right)} ^n} = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}{e^{ - \frac{{i\omega }}{2}(t - {t_0})}}\left[ {\frac{1}{{\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{x{'^2} + x'{'^2} - 2x''x'{e^{ - i\omega (t - {t_0})}}}}{{1 - {e^{ - 2i\omega (t - {t_0})}}}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{{e^{i\omega (t - {t_0})}} - {e^{ - i\omega (t - {t_0})}}}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{{e^{\frac{{i\omega }}{2}(t - {t_0})}}\sqrt {1 - {e^{ - 2i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{\sqrt {{e^{i\omega (t - {t_0})}} - {e^{ - i\omega (t - {t_0})}}} }}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}{e^{\frac{{m\omega (x{'^2} + x'{'^2})}}{{2\hbar }}}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{\hbar }\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{2i\sin (\omega (t - {t_0}))}}} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{m\omega }}{{2\hbar }}\left[ {\frac{{(x{'^2} + x'{'^2}){e^{i\omega (t - {t_0})}} - 2x''x'}}{{i\sin (\omega (t - {t_0}))}} - (x{'^2} + x'{'^2})} \right]} \right) = \\ = {\left( {\frac{{m\omega }}{{\pi \hbar }}} \right)^{1/2}}\left[ {\frac{1}{{2i\sin \omega (t - {t_0})}}} \right]\exp \left( { - \frac{{im\omega }}{{2\hbar }}\frac{{(x{'^2} + x'{'^2})\cos (\omega (t - {t_0})) - 2x''x'}}{{\sin (\omega (t - {t_0}))}}} \right)\\\end{array}
$$
The last moment you can see the Hermitian polinomial functions is when we apply your formula
