Find the mgf and E(x) of exponential family $P(X=x)=\frac{f(x)n^x}{g(n)}$ where $g(n)=\sum^\infty_{x=0}f(x)n^x$ I have to prove this is in the exponential family and find the mgf and $E(x)$

So I have that it is in the exponential family because 

$I(x)_{0,1,2,...}\frac{f(x)}{\sum^\infty_{x=0}f(x)}=h(x)$, 

$c(n)=1$,

$\sum^\infty_{x=0}exp([1-x]log(n))$ where $\sum^\infty_{x=0}exp([1-x]$ and $\sum^\infty_{x=0}exp(log(n))$ are the two parts to the exponential.

$I(x)_{0,1,2,...}\frac{f(x)}{\sum^\infty_{x=0}f(x)}*1*\sum^\infty_{x=0}exp([1-x]*log(n))$

I am not sure how to simplify the equation when taking the sum with $e^{tx}$ is put into the function.

$\sum^\infty_{x=0} e^{tx}I(x)_{0,1,2,...}\frac{f(x)}{\sum^\infty_{x=0}f(x)}*1*\sum^\infty_{x=0}exp([1-x]*log(n))$

$\sum_{x=0}^\infty \frac{f(x)(ne^t)^x}{g(n)}$

I dont know where to go from here.

When finding the mgf, does my sum on the inside cancel the sum on the top?
Note that

$$\sum_{k=0}^{\infty} f(k) n^k = \sum_{x=0}^{\infty} f(x) n^x$$



A distribution $g(x;n)$ belongs to an exponential family if it can be represented in the form:

$$g(x;n)=\nu(x)\exp[\xi(n)T(x)-S(n)]$$

We have

$$g(x;n) = P(X = x; n) = f(x) \exp[\ln(n) x -\ln[\sum_{k=0}^{\infty} f(k) n^k]]$$



$$E[X] = \sum_{j=0}^{\infty} j P(X = j; n) = \sum_{j=0}^{\infty} j[f(j) \exp[\ln(n) j -\ln[\sum_{k=0}^{\infty} f(k) n^k]]]$$

or

$$E[X] = \sum_{j=0}^{\infty} P(X \ge j; n)$$

where

$$P(X \ge j; n) =\sum_{m=j}^{\infty} f(m) \exp[\ln(n) m -\ln[\sum_{k=0}^{\infty} f(k) n^k]]$$

///ly, we have

$$E[e^{tX}] = \sum_{j=0}^{\infty} e^{tj} P(X = j; n) = \sum_{j=0}^{\infty} f(j) \exp[\ln(n) j -\ln[\sum_{k=0}^{\infty} f(k) n^k]]$$
Note that

$$\sum_{k=0}^{\infty} f(k) n^k = \sum_{x=0}^{\infty} f(x) n^x$$



A distribution $g(x;n)$ belongs to an exponential family if it can be represented in the form:

$$g(x;n)=\nu(x)\exp[\xi(n)T(x)-S(n)]$$

We have

$$g(x;n) = P(X = x; n) = f(x) \exp[\ln(n) x -\ln[\sum_{k=0}^{\infty} f(k) n^k]]$$



$$E[X] = \sum_{j=0}^{\infty} j P(X = j; n) = \sum_{j=0}^{\infty} j[f(j) \exp[\ln(n) j -\ln[\sum_{k=0}^{\infty} f(k) n^k]]]$$

or

$$E[X] = \sum_{j=0}^{\infty} P(X \ge j; n)$$

where

$$P(X \ge j; n) =\sum_{m=j}^{\infty} f(m) \exp[\ln(n) m -\ln[\sum_{k=0}^{\infty} f(k) n^k]]$$

///ly, we have

$$E[e^{tX}] = \sum_{j=0}^{\infty} e^{tj} P(X = j; n) = \sum_{j=0}^{\infty} f(j) \exp[\ln(n) j -\ln[\sum_{k=0}^{\infty} f(k) n^k]]$$