In point of fact, it was W. Ljunggren who proved in his 1942 paper "En egenskap ved de midtre binomialkoeffisienter" (Norsk. Mat. Tidsskrift 24 (1942), 18-22) that

$$p^{3n+\varepsilon} \mid F_{p}$$

where $\varepsilon=0$ if $p=2$, $\varepsilon = 1$ if $p=3$, and $\varepsilon =2$ if $p \geq 5$.

You can find the following sketch of Ljunggren's proof in pages 563-564 of the second volume of The Collected Papers of Wilhelm Ljunggren:

In the identity

$$F(y) = \prod_{\substack{\lambda =1\\ (\lambda,p)=1}}^{pm}(y-\lambda) = y^{t} -A_{1}y^{t-1} + A_{2}y^{t-2} - \ldots + A_{t-2}y^{2}-A_{t-1}y+A_{t} \quad \quad(*)$$

where $m=p^{n-1}$, $t=\varphi(p^{n})$ and $n>1$ for $p=2$, $A_{t-1}$ is divisible by $p^{\varepsilon} m^{2}$. This is a consequence of a theorem of Leudesdorf. In ($\ast$) we first put $y=pm$. Dividing by $pm$ and making use of the fact that $A_{t-1}$ is divisible by $p^{\varepsilon} \cdot m^{2}$ we easily find that $A_{t-2}$ is divisible by $mp^{\varepsilon-1}$. We then put $y=p^{2}m$ and thus get that $F(p^{2}m)-A_{t}$ is divisible by $m^{3}p^{\varepsilon+2}$. We then can give the expression [that defines $F_{p}$] in the form $$\binom{p^{2}m}{pm} - \binom{pm}{m} = \binom{pm-1}{m-1}p \frac{F(p^{2}m)-A_{t}}{A_{t}}.$$ This gives the proof of our... theorem because $(A_{t},p)=1$.

In point of fact, it was W. Ljunggren who proved in his 1942 paper "En egenskap ved de midtre binomialkoeffisienter" (Norsk. Mat. Tidsskrift 24 (1942), 18-22) that

$$p^{3n+\varepsilon} \mid F_{p}$$

where $\varepsilon=0$ if $p=2$, $\varepsilon = 1$ if $p=3$, and $\varepsilon =2$ if $p \geq 5$.

You can find the following sketch of Ljunggren's proof in pages 563-564 of the second volume of The Collected Papers of Wilhelm Ljunggren:

In the identity

$$F(y) = \prod_{\substack{\lambda =1\\ (\lambda,p)=1}}^{pm}(y-\lambda) = y^{t} -A_{1}y^{t-1} + A_{2}y^{t-2} - \ldots + A_{t-2}y^{2}-A_{t-1}y+A_{t} \quad \quad(*)$$

where $m=p^{n-1}$, $t=\varphi(p^{n})$ and $n>1$ for $p=2$, $A_{t-1}$ is divisible by $p^{\varepsilon} m^{2}$. This is a consequence of a theorem of Leudesdorf. In ($\ast$) we first put $y=pm$. Dividing by $pm$ and making use of the fact that $A_{t-1}$ is divisible by $p^{\varepsilon} \cdot m^{2}$ we easily find that $A_{t-2}$ is divisible by $mp^{\varepsilon-1}$. We then put $y=p^{2}m$ and thus get that $F(p^{2}m)-A_{t}$ is divisible by $m^{3}p^{\varepsilon+2}$. We then can give the expression [that defines $F_{p}$] in the form $$\binom{p^{2}m}{pm} - \binom{pm}{m} = p\binom{pm-1}{m-1} \frac{F(p^{2}m)-A_{t}}{A_{t}}.$$ This gives the proof of our... theorem because $(A_{t},p)=1$.

In point of fact, it was W. Ljunggren who proved in his 1942 paper "En egenskap ved de midtre binomialkoeffisienter" (Norsk. Mat. Tidsskrift 24 (1942), 18-22) that

$$p^{3n+\varepsilon} \mid F_{p}$$

where $\varepsilon=0$ if $p=2$, $\varepsilon = 1$ if $p=3$, and $\varepsilon =2$ if $p \geq 5$.

You can find the following sketch of Ljunggren's proof in pages 563-564 of the second volume of The Collected Papers of Wilhelm Ljunggren:

In the identity

$$F(y) = \prod_{\lambda =1, \, (\lambda,p)=1}^{pm}(y-\lambda) = y^{t} -A_{1}y^{t-1} + \ldots + A_{t-2}y^{2}-A_{t-1}y+A_{t} \quad \quad(*)$$

where $m=p^{n-1}$, $t=\varphi(p^{n})$ and $n>1$ for $p=2$, $A_{t-1}$ is divisible by $p^{\varepsilon} m^{2}$. This is a consequence of a theorem of Leudesdorf. In ($\ast$) we first put $y=pm$. Dividing by $pm$ and making use of the fact that $A_{t-1}$ is divisible by $p^{\varepsilon} \cdot m^{2}$ we easily find that $A_{t-2}$ is divisible by $mp^{\varepsilon-1}$. We then put $y=p^{2}m$ and thus get that $F(p^{2}m)-A_{t}$ is divisible by $m^{3}p^{\varepsilon+2}$. We then can give the expression [that defines $F_{p}$] in the form $$\binom{p^{2}m}{pm} - \binom{pm}{m} = \binom{pm-1}{m-1}p \frac{F(p^{2}m)-A_{t}}{A_{t}}.$$ This gives the proof of our... theorem because $(A_{t},p)=1$.

In point of fact, it was W. Ljunggren who proved in his 1942 paper "En egenskap ved de midtre binomialkoeffisienter" (Norsk. Mat. Tidsskrift 24 (1942), 18-22) that

$$p^{3n+\varepsilon} \mid F_{p}$$

where $\varepsilon=0$ if $p=2$, $\varepsilon = 1$ if $p=3$, and $\varepsilon =2$ if $p \geq 5$.

You can find the following sketch of Ljunggren's proof in pages 563-564 of the second volume of The Collected Papers of Wilhelm Ljunggren:

In the identity

$$F(y) = \prod_{\substack{\lambda =1\\ (\lambda,p)=1}}^{pm}(y-\lambda) = y^{t} -A_{1}y^{t-1} + A_{2}y^{t-2} - \ldots + A_{t-2}y^{2}-A_{t-1}y+A_{t} \quad \quad(*)$$

where $m=p^{n-1}$, $t=\varphi(p^{n})$ and $n>1$ for $p=2$, $A_{t-1}$ is divisible by $p^{\varepsilon} m^{2}$. This is a consequence of a theorem of Leudesdorf. In ($\ast$) we first put $y=pm$. Dividing by $pm$ and making use of the fact that $A_{t-1}$ is divisible by $p^{\varepsilon} \cdot m^{2}$ we easily find that $A_{t-2}$ is divisible by $mp^{\varepsilon-1}$. We then put $y=p^{2}m$ and thus get that $F(p^{2}m)-A_{t}$ is divisible by $m^{3}p^{\varepsilon+2}$. We then can give the expression [that defines $F_{p}$] in the form $$\binom{p^{2}m}{pm} - \binom{pm}{m} = \binom{pm-1}{m-1}p \frac{F(p^{2}m)-A_{t}}{A_{t}}.$$ This gives the proof of our... theorem because $(A_{t},p)=1$.