Multiplicity Identity in Kittel's Thermal Physics On page 25 of Kittel's Thermal Physics, the author derives the multiplicity of $N$ harmonic oscillators with total quanta of energy $n$, $g(N,n)$.
He writes
\begin{align}
g(N,n) &= \lim_{t\rightarrow 0} \frac{1}{n!}\left( \frac{d}{dt}\right)^n \sum_{s=0}^{\infty}g(N,s)t^s\\
&= \lim_{t\rightarrow 0}\frac{1}{n!}\left(\frac{d}{dt}\right)^n(1-t)^{-N}\\
&=\frac{N(N+1)(N+2)\cdots(N+n-1)}{n!}.
\end{align}
I understand everything after the first equation but I fail to see where the first equation comes from. I've tried expanding out the derivatives and summation but I still can't get it. How can I derive the first equation?
 A: I figured it out. If you pull out the summation out front, everything except the $s=n$ term vanishes. The terms with a higher power than $n$ vanish when taking the limit while the terms with a lower power than $n$ vanish when taking the $n$th derivative. 
However, it would be great if someone can come up with a better, more constructive way of deriving that formula.
A: I really like Kittel's philosophical approach to the subject in this book (counting multiplicity exactly in model systems, leveraging those systems to define entropy and temperature, ...). But here, and in many other places, his derivations/calculations seem to obfuscate rather than illuminate.
The multiplicity of $N$ quantum harmonic oscillators allowed to share $n$ energy units (sometimes called an Einstein solid),
$$g(n,N) = \frac{(n + N-1)!}{n! (N-1)!}\,,$$
is derived much more cleanly by Schroeder. He uses the "Stars and Bars" method from combinatorics, to express one particular microstate as a linear arrangement of $n$ energy units (stars) separated by $N-1$ partitions (bars), which delineate the boundaries between the $N$ individual harmonic oscillators. The multiplicity can then be seen as the total number of permutations of  $(n + N-1)$ distinguishable objects, divided by the number of permutations of stars, and by the number of permutations of bars. Alternatively (see section from Schoeder below), one can think of it as the number of ways of choosing $n$ of the $(n + N-1)$ objects to be energy units (dots).
See the relevant segment from Schroeder's text here. [Note that he uses $q$ instead of $n$ for the the number of energy units, and $\Omega$ instead of $g$ for the multiplicity.]

A: I agree with you that the derivation of the equation is quite tricky. I've also been been spending several time scratching my mind before coming to your same conclusion. 
Anyway, I would put your attention on the statement immediately before equation (51) 

To solve the problem (53) below, we need a function to represent or generate the series...

then you can think at the equation you're asking about as a trcky way for simplifying in a non-trivial way the calculations, since the coumbersome math introduced has the only goal of simplify the summation... I hope it was useful.
A: Assuming that the power series $$\sum_{s=0}^{\infty}g(N,s)t^s $$ converges for $|t|< T$ and some $T>0$, we can derive it under the sign of the series, therefore:
\begin{align}
 \frac{1}{n!}\left( \frac{d}{dt}\right)^n \sum_{s=0}^{\infty}g(N,s)t^s = 
 \sum_{s=0}^{\infty}g(N,s)\frac{1}{n!}\left( \frac{d}{dt}\right)^n t^s = 
\sum_{s=n}^{\infty}g(N,s)\frac{s(s-1)\cdots(s-n+1)}{n!} t^{s-n}
\end{align}
Since every power series uniformly converges, we can compute the limit for $t\to 0$ under the sign of series. In the last series, for $t\to 0$, all terms vanish but the one with $s=n$ giving rise to $g(N,n)$. In other words
\begin{align}
g(N,n) = \lim_{t\rightarrow 0} \frac{1}{n!}\left( \frac{d}{dt}\right)^n \sum_{s=0}^{\infty}g(N,s)t^s.
\end{align}
