Three different results for $\frac{\partial}{\partial p}\left((-\gamma+1)(v+e^\omega\frac{p^{\varepsilon+1}}{\varepsilon+1})\right)^{1/(-\gamma+1)}$

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I need to differentiate this equation wrt. $p$:

$$\frac{\partial}{\partial p}\left((-\gamma+1)\left(v+e^\omega\frac{p^{\varepsilon+1}}{\varepsilon+1}\right)\right)^{1/(-\gamma+1)}$$

See problem as image

My supervisor gets this result:

$$(-γ+1)^\frac{γ}{-γ+1}\left(v+e^{ω}\frac{p^{ε+1}}{ε+1}\right)^{\frac{γ}{-γ+1}}e^{ω}p^{ε+1}$$

See supervisor's result as image

Wolfram Alpha gets this result:

See Wolfram Alphas result as image

I get this result:

$$\left((-γ+1)e^{ω}\frac{p}{ε+1}\right)^{\frac{1}{-γ+1}-1}p^{ε+1}$$

See my result as image

Here are my calculations:

Steps: 1: Split up all parentheses (multiple inside and then add the potency of the outside parenthesis)

$$\frac{∂}{∂p}\left((-γ+1)^{\frac{1}{-γ+1}}v^{\frac{1}{-γ+1}}+(-γ+1)^{\frac{1}{-γ+1}}e^{ω^{\frac{1}{-γ+1}}}\frac{1}{ε+1}^{\frac{1}{-γ+1}} p^{\frac{ε+1}{-γ+1}}\right)$$

2: Remove $(-γ+1)^{\frac{1}{-γ+1}}v^{\frac{1}{-γ+1}}$ as it is a constant

$$\frac{∂}{∂p}\left((-γ+1)^{\frac{1}{-γ+1}}e^{ω^{\frac{1}{-γ+1}}}\frac{1}{ε+1}^{\frac{1}{-γ+1}} p^{\frac{ε+1}{-γ+1}}\right)$$

3: Use rule of derivatives (derivative wrt x: $x^n = nx^{n-1}$)

$$\left(\frac{ε+1}{-γ+1}\right)(-γ+1)^{\frac{1}{-γ+1}}e^{ω^{\frac{1}{-γ+1}}}\frac{1}{ε+1}^{\frac{1}{-γ+1}}p^{\frac{ε+1}{-γ+1}-1}$$

4: Convert the first fraction to factors by using potency rules and use the new form to multiply similar term's potences and isolate the part of the p that doesn't have the same potency in common (sorry if some terms sound strange, I didn't learn math in English)

$$(-γ+1)^{\frac{1}{-γ+1}-1}e^{ω^{\frac{1}{-γ+1}}}\frac{1}{ε+1}^{\frac{1}{-γ+1}-1}p^{\frac{1}{-γ+1}-1}p^{ε+1}$$

5: Put all with same potency under same potency

$$\left((-γ+1)e^{ω}\frac{p}{ε+1}\right)^{\frac{1}{-γ+1}-1}p^{ε+1}$$

Which one is right? What did I do wrong? Any pointers to what method I should be using?

See calculations as image

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I solved it through @KurtG. 's help. Just needed to use the chain rule:

enter image description here