Proof of Newton's Binomial Theorem I have two issues with my proof, which I will present below. Recall Newton's Binomial Theorem: 
$$(1+x)^t=1+\binom{t}{1}x+\cdot\cdot\cdot=\sum_{k=0}^\infty \binom{t}{k} x^k$$
By cleverly letting $$f(x)=\sum_{k=0}^\infty \binom{t}{k} x^k,$$
we have 
$$f'(x)=\sum_{k=1}^\infty \binom{t}{k} k x^{k-1}$$
Claim: $(1+x)f'(x)=tf(x)$ 
First problem: I would have not been able to come up with this relation had I not assumed that $f(x)=(1+x)^t$
Proof of Claim: Compare coefficients of $x^j$
For the left hand side $(1+x)f'(x)$, we have 
\begin{align*}
\binom{t}{j+1}(1+j)+\binom{t}{j}j=&j\left(\binom{t}{j}+\binom{t}{j+1}\right)+\binom{t}{j+1}\\
=&j\binom{t+1}{j+1}+\binom{t}{j+1}\\
=&\dfrac{t!t(j+1)}{(t-j)!(j+1)!}=\dfrac{t\cdot t!}{(t-j)!j!}\\
=&t\binom{t}{j}
\end{align*}
where the second inequality follows from Pascal's Triangle.
For the right hand side $tf(x)$, we have $t\binom{t}{j}$.
Second Problem: I am not quite sure if comparing coefficients is rigorous enough when proving these theorems. If I show that the degrees of the two polynomials are equal, would that be sufficient?
 A: I imagine the issue is to prove the given identity for $t\in\mathbb{R}\setminus\mathbb{N}$, otherwise induction is enough.
By definition of real exponentiation, for any $|x|<1$ we have 
$$(1+x)^t = \exp\left(t\log(1+x)\right) = \exp\left(t\sum_{k\geq 1}\frac{(-1)^{k-1}}{k}x^k\right) \tag{1}$$
hence the LHS is an analytic function in a neighbourhood of $x=0$, and
$$ (1+x)^t = 1+\sum_{m\geq 1}c_m x^m\tag{2} $$
where
$$ c_m = \frac{1}{2\pi}\int_{-\pi}^{\pi}(1+e^{i\theta})^t e^{-mi\theta}\,d\theta \tag{3} $$
by Cauchy's integral formula, but also
$$\begin{eqnarray*} c_m &=& [x^m] \left(e^{tx}\cdot e^{-tx^2/2}\cdot e^{tx^3/3}\cdots e^{(-1)^{m-1}x^m/m}\right)\\&=&(-1)^m [x^m] \left(e^{-tx}\cdot e^{-tx^2/2}\cdot e^{-tx^3/3}\cdots e^{-tx^m/m}\right)\end{eqnarray*}\tag{4} $$
due to $(1)$. It follows that
$$ c_m = \sum_{a_1+2a_2+\ldots+ma_m=m}\frac{(-t)^{a_1+a_2+\ldots+a_m}}{a_1! 2^{a_2}a_2! \cdots m^{a_m}a_m!} \tag{5} $$
and now I leave to you to understand the relation between $\binom{t}{k}$ and the RHS of $(5)$.
It may be useful to recall a few facts about exponential generating functions.

A simpler approach is to notice that both $f(x)=(1+x)^t$ and $g(x)=\sum_{k\geq 0}\binom{t}{k}x^k$ are solutions of the differential equation
$$(1+x)\cdot h'(x) = t\cdot h(x) $$
such that $h(0)=1$. $f\equiv g$ hence follows from the Cauchy-Lipschitz theorem.
A: The  combinatorial  proof  starts  from the  unique  factorization  of
permutations into disjoint cycles which gives the species
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$
with bivariate EGF $$G(z, u) = \exp\left(u\log\frac{1}{1-z}\right)$$
which yields the Stirling numbers of the first kind
$$\left[n\atop k\right]
= n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k.$$
We thus have
$$n! [z^n] G(z, u) 
= \sum_{k=1}^n \left[n\atop k\right] u^k.$$
Therefore
$$n! [z^n] G(z, u) = n! (-1)^n [z^n] G(-z, u) =
u(u+1)(u+2)\cdots(u+n-1).$$ 
Hence
$$n! (-1)^n [z^n] G(-z, -u) = (-1)^n u(u-1)(u-2)\cdots(u-(n-1))$$ 
and finally
$$n! [z^n] \exp\left(-u\log\frac{1}{1+z}\right)
\\ = n! [z^n] \exp(u\log(1+z)) = u(u-1)(u-2)\cdots(u-(n-1))$$ 
so that
$$[z^n] \exp(u\log(1+z)) = 
\frac{1}{n!} u(u-1)(u-2)\cdots(u-(n-1))
= {u\choose n}.$$
