Must n be an integer? After all, you have factorial(n-1) in there. If it must be an integer then just test a list of integer values for n. That will not be a long list, as you will have massive numerical problems if n is large. Pick the value of n that makes you happiest.
Hint: no, if n is an integer, you will never find an integer value for n that will come even remorely close to that data.
Instead, you very much need to use gamma(n) in there, instead of factorial(n-1). They are the same for integers, but the gamma function extends factorial nicely onto the real line.
Next, just to be safe, I used realpow in this. Although the .^ operator should be fine here.
Tau = 5.2
E = [0 0.0199867 0.0999334 0.1598934 0.1998668 0.1598934 0.1199201 0.0799467 0.05996 0.0439707 0.02998 0.011992 0]';
t = [0 1 2 3 4 5 6 7 8 9 10 12 14]';
E_t = @(t,n,tau) realpow(n,n).*realpow(t,n-1).*exp(-n*t/tau)./gamma(n)./realpow(tau,n);
Having done that, it is worth testing your fucntion to see if it has a chance to fit your data. So I played around with a few arbitrary values for n, and 4.25 seems to work resonably well, except that it does not reach the peak.
fplot(@(t) E_t(t,4.25,5.2),[0,14]),hold on,plot(t,E,'o')
As you see, it fits all ot the points pretty well, but it misses that peak. However, if you increase n to the point where is is large enough to hit the peak, then it looks like utter crap for the rest of the data.
fplot(@(t) E_t(t,6,5.2),[0,14]),hold on,plot(t,E,'o')
Could I have used an optimization to find the best value of n? Well, yes. That would be easy now. But the best value of n will be a compromise, likely close to 4.25, as I found initially.
fun = @(n) norm(E - E_t(t,n,tau));
[nopt,fval,exitflag] = fminsearch(fun,5)
fplot(@(t) E_t(t,nopt,5.2),[0,14]),hold on,plot(t,E,'o')
This points out several things. First, that I have a good eye for getting the best fit, even without a computer. But it also suggests that your data does not fit that model perfectly, missing that peak, even when the rest of the data fots nicely..