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It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969):

Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions, 19 (10). pp. 709-712. https://dx.doi.org/10.1080/10652460802332342

The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form:

\begin{equation*} W e^W = x \implies W'/W + W' = 0 \implies \{W Y - x=0, \frac{Y'}{Y} = W'\} . \end{equation*}

Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$, $Y'/Y = W'$ and $f(x,Y,W)=0$, with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$. According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$. So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it would be algebraic.

It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.

It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969):

Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions, 19 (10). pp. 709-712. https://dx.doi.org/10.1080/10652460802332342

The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form:

\begin{equation*} W e^W = x \implies W'\!/W + W' = 1/x \implies \{Y = x/W,\; Y'\!/Y = W'\} . \end{equation*}

Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$, $Y'/Y = W'$ and $f(x,Y,W)=0$, with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$. According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$. So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it would be algebraic.

It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.

It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969):

Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions, 19 (10). pp. 709-712. http://dx.doi.org/10.1080/10652460802332342

The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form:

\begin{equation*} W e^W = x \implies W'/W + W' = 0 \implies \{W Y - x=0, \frac{Y'}{Y} = W'\} . \end{equation*}

Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$, $Y'/Y = W'$ and $f(x,Y,W)=0$, with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$. According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$. So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it would be algebraic.

It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.

It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969):

Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions, 19 (10). pp. 709-712. https://dx.doi.org/10.1080/10652460802332342

The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form:

\begin{equation*} W e^W = x \implies W'/W + W' = 0 \implies \{W Y - x=0, \frac{Y'}{Y} = W'\} . \end{equation*}

Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$, $Y'/Y = W'$ and $f(x,Y,W)=0$, with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$. According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$. So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it would be algebraic.

It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.

It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969):

Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions, 19 (10). pp. 709-712. http://dx.doi.org/10.1080/10652460802332342

The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form:

\begin{equation*} W e^W = x \implies W'/W + W' = 1/x \implies \{W Y - x=0, \frac{Y'}{Y} = W'\} . \end{equation*}

Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$, $Y'/Y = W'$ and $f(x,Y,W)=0$, with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$. According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$. So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it is algebraic.

It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.

It seems that the non-linear differential equation satisfied by the Lambert $W$ function is simple enough for this question to have already been answered. This paper proves that the Lambert $W$ is non-elementary by appealing to a result of Rosenlicht (1969):

Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J. (2008) Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions, 19 (10). pp. 709-712. http://dx.doi.org/10.1080/10652460802332342

The argument runs roughly as follows. First note that $W(x)$ satisfies a two-variable algebraic-differential system, where the differential part has a special form:

\begin{equation*} W e^W = x \implies W'/W + W' = 0 \implies \{W Y - x=0, \frac{Y'}{Y} = W'\} . \end{equation*}

Now, consider the differential field $\mathbb{C}(x)(Y,W)$ where $x'=1$, $Y'/Y = W'$ and $f(x,Y,W)=0$, with $f$ polynomial in $Y$ and $W$ over $\mathbb{C}(x)$. According to a more general result of Rosenlicht, the field $\mathbb{C}(x)(Y,W)$ is Liouvillian only if $Y$ and $W$ are algebraic over $\mathbb{C}(x)$. So, if $W(x)$ were elementary (elementary is a special case of Liouvillian) then it would be algebraic.

It remains to check that $W(x)$ is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.