Negative binomial maximum likelihood The pdf of a negative binomial is 
$$θ(X=x)= \left( \begin{array}{c}
x+j-1  \\
x  \end{array} \right)(1-θ)^xθ^j,$$
How would I create the likelihood of this function in order to maximize θ?And how does the likelihood change if there is n observations vs. 1 observation?
So far, I have that the likelihood is 
∏ (j + x − 1 C x)θ^j(1-θ)^x
which is simplified to (through the derivative of the log-likelihood)
=jnlnθ+∑Xln(1-θ)
Meaning that jn/θ=∑X/1-θ
I solved for theta and have:
θ-hat= jn/(jn+∑x)
Is there something wrong? If so, where?

Something extra:
(The alternative pdf is $$P(X=x|j,θ)= \binom{x-1}{j-1}θ^{j}(1-θ)^{x-j}$$
would it yield the same likelihood?)

You are using the PDF
$$P(X = x|j, \theta)
= {x + j - 1 \choose x}\theta^j(1-\theta)^{x},$$
for $x = 0, 1, \dots$, where $x$ is the number of
failures observed before seeing the $j$th success
and $\theta$ is the probability of success on any
one trial. I am being fussy about writing everything
out in full, because (as you will see) every detail
is important here.
With data $X_1, X_2, \dots, X_n,$
you set the partial derivative of the loglikelihood 
function equal to $0$ to find the estimator
$$\hat \theta = \frac{jn}{jn + T},$$
where $T = \sum_{i=1}^n x_i,$ which is the total number
of failures seen while waiting for $nj$ successes.
If I read the notes on your derivation correctly, I think
you did it correctly. 
Furthermore, this estimator of $\theta$ makes sense,
because the numerator is the number of successes observed
and the denominator is the total number of trials
performed.

Now to try to clear up some of your additional questions
in the original post and arising from Comments.
(1) What changes if $n = 1$? There is no product sign
in the likelihood and so summation sign in the equation to
solve for the estimator. The estimator becomes 
$\hat \theta = j/(j + x).$ Again the estimator is the
ratio of the number $j$ of successes to the total number
of trials until those $j$ successes are observed.
2) If you change to the alternative PDF, the estimator
will still be the number $nj$ of success divided by the total
number of trials necessary to to observe them. However, the
equation will not look the same, because this is the PDF
of a different random variable. Now $X$ is defined as the
total number of $trials$ (not $failures$) until $nj$
successes are seen. This random variable takes values
$x = j,\, j+1,\, j+2,\, \dots.$
3) Your answer is not the same as the one in Wikipedia
because the random variable there counts the number of
$successes$ until (in your notation) $nj$ failures are
observed. (I frequently refer questioners to Wikipedia,
but not always. Sometimes is is not at the right level,
has confusing notation, or formulates an issue in
a different way than in the problem posted. Sometimes Wikipedia
is helpful, sometimes not; here I think not.)
Please check everything here. I am not the world's
best proofreader. And leave a Comment for me if there
are unresolved issues about your question. I, or
someone else, will try to help.
