Solution:

We show $P(x)<0$ for all $x>0$, i.e. $M(x+1)^{k}<\left(x+a_{1}\right) \ldots\left(x+a_{n}\right) \Leftrightarrow$ $a_{1} a_{2} \ldots a_{n}(x+1)^{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}}<\left(x+a_{1}\right) \ldots\left(x+a_{n}\right) \Leftrightarrow \prod_{i=1}^{n} a_{i}(x+1)^{\frac{1}{a_{i}}}<\prod_{i=1}^{n}\left(x+a_{i}\right)$.

For this we show that for every $i\left((1 \leq i \leq n)\right.$ the relation $a_{i}(x+1)^{\frac{1}{a_{i}}} \leq x+a_{i}(1)$ holds, and that for at least one $i$ we even have $a_{i}(x+1)^{\frac{1}{a_{i}}}<x+a_{i}$. The claim then follows by multiplying over all $i$.

From the AM-GM inequality for the numbers $x+1,1,1, \ldots, 1\left(a_{i}-1\right.$ summands 1$)$ it follows that $\frac{x+a_{i}}{a_{i}} \geq \sqrt[a]{x+1}$, which after multiplying by $a_{i}$ yields exactly (1). Equality holds precisely for $a_{i}=1$, and this cannot hold for all $i$, since then $M=1$ would follow, contradicting the assumption $M>1$. Since for the given values all transformations are permissible, everything is proved.