Question: How to prove the following?

$$\frac{d^a{(x}^a)}{dx^a}+\frac{d^{a-1}{(x}^a)}{dx^{a-1}}+\frac{d^{a-2}{(x}^a)}{dx^{a-2}}+\dots +\frac{d^2{(x}^a)}{dx^2}+\frac{d^1{(x}^a)}{dx^1}+\frac{d^0{(x}^a)}{dx^0}+\int^x\_0{{(x}^a)dx}+\int^x\_0{\int^x\_0{{(x}^a)dx}dx}+\int^x\_0{\int^x\_0{\int^x\_0{{(x}^a)dx}}dxdx}+\dots =a!e^x $$

where, $a$ is a constant.

Answer: 
Your expression conatisn a finite sum of derivatives and an infinite sum of integrals of $x^a$. Let's look at them separately.

Each derivative lowers the power by one and provides a factor of the power in front, so we have 
$$ \frac{\text{d}^p}{\text{d}x^p} x^a=a\cdot(a-1)\dotsm(a-p+1) \cdot x^{a-p}\\
=a\cdot(a-1)\dotsm(a-p+1)\cdot\frac{(a-p)!}{(a-p)!} \cdot x^{a-p}\\
=a!\cdot\frac{x^{a-p}}{(a-p)!}\,.$$
Hence, we recognise the terms in the series for $e^x$, multiplied with $a!$ -- the derivatives give the series up to the $a$-th term.

The integrals provide the rest. To see this, look at e.g. the two-integral term, which should be written as
$$I\_2=\int\_0^x\left(\int\_0^{x\_2} x\_1^a \,\text{d} x\_1 \right)\text{d}x\_2$$
to get the limits and integration variables right. The integrations give
$$I\_2=\int\_0^x\frac{a+1} x\_2^{a+1} \,\text{d}x\_2 = \frac{1}{(a+1)(a+2)}\, x^{a+2}\\
=\frac{a!}{a!}\cdot\frac{1}{(a+1)(a+2)}\,x^{a+2}\\
=a! \cdot \frac{x^{a+2}}{(a+2)!} \,.$$
Again, we see a term in the series for $e^x$, and the rest of the proof is straightforward.

Effectively, your original expression takes the series expansion of $e^x$, splits it into terms with powers up to $a$ and larger than $a$ and expresses these as derivatives/integrals of a monomial.
$$a!\;e^x=a!\sum\limits\_{n=0}^{\infty}\frac{x^n}{n!}=\sum\limits\_{n=-a}^{\infty}\frac{a!\;x^{a+n}}{(a+n)!}=\sum\limits\_{n=-a}^{\infty}\frac{d^{-n}(x^a)}{dx^{-n}}$$

$$=\sum\limits\_{n=-a}^{0}\frac{d^{-n}(x^a)}{dx^{-n}}+\sum\limits\_{n=1}^{\infty}\frac{d^{-n}(x^a)}{dx^{-n}}=\sum\limits\_{n=0}^{a}\frac{d^{n}(x^a)}{dx^{n}}+\sum\limits\_{n=1}^{\infty}\frac{d^{-n}(x^a)}{dx^{-n}}$$