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I don't know about gamma function, but if I were to extend the definition of factorials in an intuitive and natural way, I would do it like this:

Suppose we want to get the value of $5.5!$.

So, I need to get in the middle of $5!$ and $6!$ intuitively.

To get to $6!$ from $5!$, we multiply $5!$ by 6, i.e. we apply the function $f(x)=6x$ to $x=5!$.

Since, we have to get in the middle of this operation, I'd apply the funcional-square root of $f(x)=6x$, i.e. $\sqrt{6}x$ to $5!$ so, $5.5!=5!*\sqrt{6}$ by this definition.

Similarly, To get $7.1!$ I would apply the functional-tenth root of $f(x)=8x$, i.e., $f(x)=x*8^{0.1}$ to 7! which gives $7!*8^{0.1}$.

So, my extension would be: To get $x!$: If $k$ is the fractional-part of $x$ and $a$ is its integer part, then $$x!=a!\cdot (a+1)^k$$

I couldn't understand much about gamma function, but I can surely say one thing that this one is a lot simpler. Is there anything wrong with this?

UPDATE: I did some calculations on my calculator and found that my factorial definition gives values relatively close to the gamma function.

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    What would be for example $\pi !$ in your definition? – 2017-02-21
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    @TheGeekGreek It would be $3! \cdot 4^{\pi-3}$; they gave the definition. – 2017-02-21
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    @TheGeekGreek Do you have any trouble in putting $3!*4^{\pi-3}$ in your calculator? I never said that the fractional part can't be a never-ending number. Did you downvote it? – 2017-02-21
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    I think this is an interesting question, but you might get a better response by asking why the Gamma function is considered a 'natural' extension of the factorial, and why yours doesn't fit those criteria – 2017-02-21
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    @Dove An important property of the factorial function is that $x \cdot (x-1)! = x!$. For your definition of the factorial function, is it true that $4.3! = 4.3 \times 3.3!$? – 2017-02-21
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    @user49640 Ok. I think that's a problem. In my definition that'd only be true for integer values. I understand that that's a problem. – 2017-02-21
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    @Dove If you think about it, you'll see that the formula $x! = x \cdot (x-1)!$ really isn't enough on its own. You could define the factorial function practically any way you like between $1$ and $2$, and you'll still be able to extend it to all positive $x$ in such a way that the formula holds true. What you need, in addition to that, is to note that if $x$ is an integer, then the limit of $\frac{(n+x)!}{n!n^x}$ as $n$ tends to infinity is always $1$. If you want both this formula and the last one to hold true for all real values of $x$, *then* there will be only one function that works. – 2017-02-21
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    @user49640 So, is the gamma function the only possible extension? And, I guess the first requirement makes the function have a great property. But why the second requirement? – 2017-02-21
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    There are several alternative “gamma functions”. See here, for example: http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunctionMJ.html – 2017-02-21
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    @Dove I mean the gamma function is the only possible extension satisfying both conditions. Do you mean why would we demand that the second requirement be satisfied? Well, it's true for integer values of $x$, so it's not unreasonable to expect it to remain true for real $x$. That being said, it could be argued to be a matter of taste. It will probably seem more natural to you if you carry out the proof that it's true when $x$ is an integer. – 2017-02-21

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See https://proofwiki.org/wiki/Gamma_Function_is_Unique_Extension_of_Factorial

Essentially, once you want alternative extension of factorial, you lose either factorial-like behaviour, or lose log-convexity, and both are very commonly used property of factorial (or you lose analytic, which is probably even worse).

Also from number theory point of view, Gamma function is much more useful due to L-function.