If `P_m` stands for `m_{P_{m}}` , then prove that: `1+1. P_1+2. P_2+3. P_3++ndotP_n=(n+1)!`

Explanation

To prove the equation:

\[ 1 + 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 + \ldots + n \cdot P_n = (n + 1)! \]

where \( P_m \) stands for \( m \cdot P_{m-1} \) and \( P_0 = 1 \), we will follow these steps:

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