Are these two expressions equivalent? Checking by hand or Mathematica

Question

Are the following two expressions equivalent?

Expression 1: $$\sum_{k=0}^{n-1}\dfrac{x^k}{k!}\sum_{L=0}^{\infty}\dfrac{(k+L)!}{L!}y^L$$

Expression 2: $$\sum_{L=0}^{\infty}\dfrac{y^L}{L!}\sum_{k=0}^{n-1}\dfrac{(k+L)!}{k!}x^k$$

WolframAlpha doesn't think so, but I am not convinced:

Expression 1: http://www.wolframalpha.com/input/?i=%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cdfrac%7Bx%5Ek%7D%7Bk!%7D%5Csum_%7BL%3D0%7D%5E%7B%5Cinfty%7D%5Cdfrac%7B(k%2BL)!%7D%7BL!%7Dy%5EL

Expression 2: http://www.wolframalpha.com/input/?i=%5Csum_%7BL%3D0%7D%5E%7B%5Cinfty%7D%5Cdfrac%7By%5EL%7D%7BL!%7D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%5Cdfrac%7B(k%2BL)!%7D%7Bk!%7Dx%5Ek

How can I check this directly in Mathematica?

Answer 1

$$\sum_{k=0}^{n-1}{x^k\over k!}\sum_{L=0}^{\infty}{(k+L)!\over L!}y^L =\sum_{k=0}^{n-1}\sum_{L=0}^{\infty}{x^k\over k!}{(k+L)!\over L!}y^L =\sum_{L=0}^{\infty}\sum_{k=0}^{n-1}{x^k\over k!}{(k+L)!\over L!}y^L =\sum_{L=0}^{\infty}{y^L\over L!}\sum_{k=0}^{n-1}{(k+L)!\over k!}x^k$$

Answer 2

I have no knowledge of Mathematica, but as an aside I'll point out that the first version allows easy calculation of a closed form:

$$\begin{align*}\sum_{k=0}^{n-1}\frac{x^k}{k!}\sum_{L\ge 0}\frac{(k+L)!}{L!}y^L&=\sum_{k=0}^{n-1}\frac{x^k}{k!}\sum_{L\ge 0}\binom{k+L}kk!y^L\\ &=\sum_{k=0}^{n-1}x^k\sum_{L\ge 0}\binom{k+L}ky^L\\ &=\sum_{k=0}^{n-1}\frac{x^k}{(1-y)^{k+1}}\\ &=\frac{1}{1-y}\cdot\frac{\frac{x^n}{(1-y)^n}-1}{\frac{x}{1-y}-1}\\ &=\frac{x^n-(1-y)^n}{(x+y-1)(1-y)^n}\;. \end{align*}$$

For the second I see nothing better than reversing the order of summation.